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CLEC'TIC  EDUCATIONAL  HEEJJiS. 


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NEW  PRACTICAL 


ARITHMETIC 


A  Revised  Edition   of  the   Practical  Arithmetic 


JO^^'E'PH"  KA'Y/K/R-,:     * 

Ijjie  Professor  in    Woodward   College. 


YAN   ANTWERP,   BRAGG   &   CO. 


137WALNUT  Street, 
CINCINNATI. 


28   Bond   Street 
NEW  YORK. 


RAY'S   MATHEMATICAL  SERIES,     ^f^^f^ 

^_ ^e/^V, 

Ray'^s  Neiv  Primary  Arithmelic. 

Ray^s  New  Intellectual  Arithinetico 

Rafs  New  Practical  Aritlimetic. 

Ray^s  New  Elementary  Arithmetic. 

Rafs  Higher  Arithmetic, 

Ray's   Test  Examples. 

Ray^s  Neiv  Elemeittary  Algeh'a. 

Rays  New  Higher  Algebra. 

Ray^s  Plane  and  Solid  Geometry. 

Ray^s   Geojnetry  and  Trigonouietry. 

Rafs  Analytic   Geometry>. 

Ray^s  Elements  of  Astronomy. 

Ray's  Surveying  and  Navigatian. 

Ray's  D>j^'er2ntial  i0^d\l\ii^gral\  Calculus. 


Copyright 

1877 

BY  Van  Antwerp,  Bragg  &  Co. 

EDUCATION  OEfY* 


ECLECTIC  PRESS, 

VAN  ANTWERP,  BRAGG  &  CO. 

CINCINNATI. 


yti 


tM{ 


PREFACE. 


Changes  in  the  methods  of  instruction  in  our  schools  and  in  the 
modes  of  transacting  business  have  made  it  necessary  to  revise 
Ray's  Practical  Arithmetic. 

No  other  work  on  Arithmetic  ever  had  so  extensive  use  or  wide- 
spread popularity.  Teachers  every-where,  throughout  the  length 
and  breadth  of  the  land,  are  familiar  with  its  pages,  and  millions  of 
pupils  have  gained  their  arithmetical  knowledge  from  the  study  of 
its  principles.  More  than  ten  thousand  editions  of  it  have  gone 
forth  from  the  press. 

In  view  of  these  facts,  it  has  been  the  constant  aim  in  making  this 
revision  to  preserve  carefully  those  distinctive  features  of  the  former 
editions,  which  constituted  the  peculiar  philosophical  method  of  its 
learned  author,  viz.: 

1st.  Every  principle  is  clearly  explained  by  an  analysis  or  solu- 
tion of  simple  examples,  from  which  a  Rule  is  derived.  This  is 
followed  by  graduated  exercises  designed  to  render  the  pupil  familiar 
with  its  application. 

2d.  The  arrangement  is  strictly  philosophical;  no  principle  is 
anticipated;  the  pupil  is  never  required  to  perform  any  operation 
until  the  principle  on  which  it  is  founded  has  first  been  explained. 

The  changes  made  fall  naturally  under  two  heads:  (1)  those  which 
adapt  the  book  better  to  the  advanced  methods  of  instruction;  (2) 
those  which  exhibit  present  methods  of  computation  in  business. 

In  the  first  place,  special  attention  is  invited  to  the  beauty  and 
elegance  of  the  typogranhy^^^  The^^ffiyant  matter  of  the  volume, 

961b;>y  (iii) 


IV  PKEFACE. 

the  definition,  the  solution,  or  the  rule,  is  at  once  clearly  indicated 
by  a  difference  of  type.  A  running  series  of  articles,  with  numbered 
paragraphs,  enhances  the  convenience  of  the  text-book  for  recitation 
and  for  reference. 

The  analytic  solutions  and  written  operations  have  been  carefully 
separated.  All  obsolete  Tables  of  Weights  and  Measures,  such  as 
Beer  Measure  and  Cloth  Measure,  and  all  obsolete  denominations, 
such  as  drams,  roods,  etc.,  are  discarded.  The  Metric  System  of 
Weights  and  Measures  is  presented  in  accordance  with  its  now 
widely  extended  usage,  and  is  assigned  its  proper  place  immediately 
after  Decimals. 

A  few  subjects,  such  as  Factoring  and  the  principles  of  Frac- 
tions, have  been  entirely  rewritten,  and  in  many  instances  the 
definitions  and  rules  have  been  simplified.  The  subject  of  Percent- 
age has  been  much  expanded,  and  an  endeavor  has  been  made  to 
systematize  its  numerous  applications;  many  novel  and  interesting 
features,  both  of  subject-matter  and  classification,  will  here  be  met 
with  for  the  first  time.  The  subjects  of  Interest  and  Discount 
have  received  that  careful  attention  which  their  importance  demands. 

The  publishers  desire  to  express  their  thanks  to  the  many 
teachers  whose  suggestions  and  corrections  are  embodied  in  the 
present  edition.  Especial  mention  is  due  Prof,  M.  W.  Smith  and 
Mr.  A.  P.  Morgan  for  many  valuable  features  of  this  revision. 

In  conclusion,  the  publishers  wish  to  reiterate  that  the  object 
throughout  has  been  to  combine  practical  utility  with  scientific 
accuracy;  to  present  a  work  embracing  the  best  methods,  with  all 
real  improvements.  How  far  this  object  has  been  secured  is 
again  submitted  to  those  engaged  in  the  laborious  and  responsible 
work  of  education. 

Cincinnati,  August,  1877. 


TABLE   OF   CONTENTS. 


PAGE 

The  Arabic  System  of  Notation 9 

The  Koman  System  oe  Notation 20 

Addition 22 

Subtraction 31 

Multiplication 39 

Contractions  in  Multiplication 47 

Division 50 

Short  Division  .         . ,       .  54 

Long  Division 59 

Contractions  in  Division .  64 

General  Principles  of  Division 67 

Compound  Numbers 71 

United  States  Money 72 

Merchants'  Bills 83 

Reduction  of  Compound  Numbers 84 

Dry  Measure 84 

Rules  for  Reduction  .         . 87 

Liquid  Measure 88 

Avoirdupois  Weight 89 

Long  Measure 90 

Square  Measure 90 

Solid  or  Cubic  Measure 94 

Time  Measure 96 

(V) 


Vi  CONTENTS. 

PAGM 

Miscellaneous  Tables 97 

Addition  of  Compound  Numbers      .         .         ,         .         .  102 

Subtraction  of  Compound  Numbers          ....  106 

Multiplication  of  Compound  Numbers     ....  Ill 

Division  of  Compound  Numbers 113 

Longitude  and  Time 115 

Factoring 118 

To  Find  the  Greatest  Common  Divisor    ....  123 

To  Find  the  Least  Common  Multiple      ....  125 

Cancellation 127 

Fractions 131 

Principles 135 

Reduction  of  Fractions 137 

Addition  of  Fractions 144 

Subtraction  of  Fractions 147 

Multiplication  of  Fractions       .         .         .        .         .         .  149 

.    Compound  Fractions 152 

Division  of  Fractions 154 

Complex  Fractions 157 

Fractional  Compound  Numbers 159 

Practice 165 

Decimal  Fractions 168 

Reduction  of  Decimals 175 

Addition  of  Decimals 178 

Subtraction  of  Decimals 179 

Multiplication  of  Decimals 180 

Division  of  Decimals         .         .         .         .         .         .         .  183 

Decimal  Compound  Numbers  .         .         .         .         .         .  186 

The  Metric  System 189 

Measures  of  Length 190 

Land  or  Square  Measure 192 

Measures  of  Capacity 192 

Measures  of  Weight 193 

Table  of  Values         ........  194 


CONTENTS.  vii 

PAGK 

Percentage 197 

Formulas  for  the  four  cases  of  Percentage.      .         .         .  203 

Applications  of  Percentage 205 

Mercantile  Transactions 206 

Commission 206 

Trade  Discount 208 

Profit  and  Loss 210 

Stock  Transactions 213 

Brokerage 214 

Assessments  and  Dividends      .         .         .         .         .         .215 

Stock  Values 216 

Stock  Investments 217 

Interest 219 

Simple  Interest 221 

The  Twelve  Per  Cent  Method 229 

Formulas  for  the  five  cases  of  Interest     .         .         .         .  237 

Compound  Interest 237 

Annual  Interest 239 

Partial  Payments 241 

Discount 247 

Bank  Discount 247 

True  Discount  . .256 

Exchange 260 

Domestic  Exchange 261 

Foreign  Exchange 262 

English  Money 262 

French  Money '        .         .  263 

German  Money 263 

Canadian  Money 263 

Insurance 265 

Fire  and  Marine  Insurance 265 

Life  Insurance .         .  267 


viii  CONTENTS. 

PAGK 

Taxes 269 

State  and  Local  Taxes 269 

United  States  Revenue 273 

Internal  Revenue      .         .  274 

Duties  or  Customs     ....  ...  274 

Ratio .276 

Principles 280 

Proportion 282 

Simple  Proportion 285 

Compound  Proportion       .         .         .         .         .         .         .  289 

Partnership 291 

Bankruptcy 293 

General  Average "        .         .         .  293 

Partnership  with  Time 294 

Equation  of  Payments 295 

Average 297 

Involution         , 298 

Evolution .        .        .  300 

Square  Root 302 

Cube  Root 309 

Mensuration 316 

Measurement  of  Surfaces 316 

Measurement  of  Solids 323 

Applications  of  Mensuration 328 

Progressions 331 

Arithmetical  Progression 331 

Geometrical  Progression 334 


Article  1.'  1.  A  Unit  is  a  single  thing  of  any  kind; 
as,  one,  one  apple,  one  dollar,  one  pound. 

2.  A  Number  consists  of  one  or  more  units ;  as,  one, 
five,  seven  cents,  nine  men. 

3.  Arithmetic  treats  of  nimibers,  and  is  the  art  of 
computing  by  them. 

4.  Numbers  are  expressed  in  two  ways ;  first,  by  words; 
second,  by  characters. 

5.  A  System  of  Notation  is  a  method  of  expressing 
numbers  by  characters. 

6.  Two  systems  of  Notation  are  in  use,  the  Arabic 
and. the  Roman.  The  Arabic  system  is  used  in  all  our 
arithmetical  calculations. 


THE  ARABIC  SYSTEM  OF  NOTATION. 

2,  1.  To  express  numbers,  the  Arabic  Notation  em- 
ploys ten  characters,  called  figures;  namely,  1,  2,  3,  4,  5, 
6,  7,  8,  9,  0. 

Eemark  1. — The  Arabic  System  of  N'otation  is  so  called  because 
its  characters  appear  to  have  been  introduced  into   Europe   bj'  the 

(9) 


10 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Arabians;  tut  i'c  is  now  o-cnv^r.illy  acknowledged  that  they  originated 
in  India. 

Rem.  2^; — ^T'ae 'A^f'J^hia  Not'ition  is  also  called  the  Decimal  System 
and  the  ^ClnrvrTio'ii  '%,s^m.   '   ' 

2.  The  Order  of  a  figure  is  the  place  it  occupies  in  a 
number. 


UNITS  OF  THE  FIRST  ORDER,  OR  UNITS. 


3. 


1.  A   unit    or    single    thing    is  owe, 
One      unit    and  one  more  are  two^ 
Two     units  and  one  more  are  three^ 
Three  units  and  one  more  are  four^ 
Four    units  and  one  more  are  five. 
Pive     units  and  one  more  are  six^ 
Six       units  and  one  more  are  seven. 
Seven  units  and  one  more  are  eight, 
Eight  units  and  one  more  are  nine, 


written  1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 


2.  These  nine  characters  are  called  significant  figures, 
because  they  denote  something. 

3.  The  character  0,  called  naught,  stands  for  nothing; 
its  use  is  to  fill  vacant  orders.  The  0  is  also  called 
cipher  and  zero. 

4.  When  a  figure  stands  alone  or  in  the  first  place  at 
the  right  of  a  number,  it  represents  one  or  more  units 
of  the  first  order. 

5.  Units  of  the  first  order  are  called  simply  units; 
and  the  place  they  occupy  is  called  the  units'  place. 


UNITS  OF  THE  SECOND  ORDER,  OR  TENS. 

4.     1.  Nine    units    and    one    more   are    called    ten;    it 
Iso    is    represented    by    the    figure    1 ;    but    the    one    is 


NOTATION. 


11 


made    to    occupy  the    second    place    from    the    right    by 
writing  a  0  in  the  units'  phice. 


2.  One  ten  is  written  thus  .  . 
Two  tens  are  twenty^  written 
Three  tens  are  thirty,  '^ 

Four    tens  are  forty,  " 

Five     tens  are  fifty,  '' 

Six       tens  are  sixty,  " 

Seven  tens  are  seventy,        '' 
Eight  tens  are  eighty,  " 

Nine    tens  are  ninety,  ^' 


10. 
20. 
30. 
40. 
50. 
60. 
70. 
80. 
90. 


3.  When  a  figure  in  a  number  stands  in  the  second 
place  from  the  right,  it  represents  one  or  more  units  of 
the  second  order. 

4.  Units  of  the  second  order  are  called  tens;  and  the 
place  they  occupy  is  called  the  tens'  place. 


TENS   AND   UNITS.  , 

5.  1.  The  numbers  between  10  and  20,  20  and  30, 
etc.,  are  expressed  by  representing  the  tens  and  units 
of  which  they  are  composed. 

2.  One  ten    and  one      unit    are  eleven,  written  11. 

and  two      units  are  twelve,  "        12. 

and  three  units  are  thirteen,  ^'        13. 

and  four     units  are  fourteen,  "        14. 

and  five      units  are  fifteen,  "        15. 

and  six       units  are  sixteen,  "        16. 

and  seven  units  are  seventeeti,  "        17. 

and  eight   units  are  eighteen,  "        18. 

and  nine     units  are  nineteen,  '•        19. 

unit    are  twenty -one,  ''        21. 

units  are  twenty -two,  ''        22. 


One  ten 

One  ten 

One  ten 

One  ten 

One  ten 

One  ten 

One  ten 

One  ten 

Two  tens  and  one 

Two  tens  and  two 


12 


KAY'S    NEW    PK ACTIO AL   AKITHMETIC. 


]S^UMBERS   TO   BE  WRITTEN. 

1.  Twenty-three;     twenty-four;     twenty-five;    twenty- 
six;    twenty-seven;    twenty -eight ;    twenty- nine. 

2.  Thirty-seven  ;    forty-tw^o ;    fifty-six ;    sixty-nine ;  sev- 
enty-three ;  eighty-seven ;  ninety-four. 

3.  Eighty-three;  forty-five;  ninety-nine;  fifty-one;  thir- 
ty-six: seventy-eight;  sixty -two. 

4.  Fifty-five ;    ninety -three  ;     eighty-one ;     sixty-seven  ; 
forty-nine ;  seventy-four ;  thirty-eight. 

5.  Seventy-six ;     forty-four ;     eighty -two ;     fifty-seven  ; 
thirty-five ;  ninety-one ;  sixty-three. 


NUMBERS   TO   BE   READ. 


1. 

71; 

32 

53 

84 

65 

46; 

2. 

58; 

34 

79 

66 

41 

85; 

3. 

75; 

43 

88 

61 

59 

33; 

4. 

39; 

72 

54 

86 

47 

98; 

5. 

6S; 

77 

31 

89 

52 

96; 

97. 
92. 
95. 
64. 

48. 


UNITS  OF  THE  THIRD  ORDER,  OR  HUNDREDS. 

6.  1.  Ten  tens  are  one  hundred;  it  is  represented  by 
the  figure  1  written  in  the  third  order,  the  orders  of 
tens  and  units  being  each  filled  with  a  cipher. 


One      hundred  is  wa^itten 

thus, 

100 

Two     hundred    "         " 

200. 

Three  hundred    *v        '' 

300 

Four    hundred    ''         " 

400 

Five     hundred    ''         " 

500 

Six       hundred    ''         " 

600. 

Seven  hundred    ''         " 

700. 

Eight   hundred    "         " 

800. 

Nine     hundred    ''         "    . 

900. 

NOTATION.  13 

2.  Units  of  the  third  order  are  called  hundreds;  and 
the  place  they  occuj^y  is  called  the  hundreds'  place. 

HUNDREDS,  TENS,  AND  UNITS. 

7.  1.  The  numbers  between  100  and  200,  200  and 
300,  etc.,  are  expressed  by  representing  the  hundreds, 
tens,  and  units  of  which  they  are  composed. 

2.  One  hundred  and  one  unit  are  one  hundred  and  one, 
written  101. 

One  hundred  and  one  ten  are  one  hundred  and  ten, 
written  110. 

One  hundred  and  one  ten  and  one  unit  are  one  hun- 
dred and  eleven,  written  111. 

One  hundred  and  two  tens  are  one  hundred  and  twenty, 
written  120. 

One  hundred,  two  tens,  and  five  units  are  one  hundred 
and  twenty -five,  written  125. 

NUMBERS   TO  BE  WRITTEN. 

1.  One  hundred  and  thirty;  one  hundred  and  forty; 
one  hundred  and  fifty ;  one  hundred  and  sixty ;  one  hun- 
dred and  seventy;  one  hundred  and  eighty. 

2.  One  hundred  and  twenty-three ;  four  hundred  and 
fifty-six ;  seven  hundred  and  eighty-nine ;  one  hundred 
and  forty-seven ;  two  hundred  and  fifty-eight ;  three 
hundred  and  sixty -nine. 

3.  One  liundred  and  two;  three  hundred  and  forty- 
five  ;  six  hundred  and  seventy-eight ;  two  hundred  and 
thirty-four;  five  hundred  and  sixty-seven;  eight  hundred 
and  ninety. 

4.  Four  hundred  and  fifty -three  ;  seven  hundred  and 
eighty-six;  nine  hundred  and  twelve;  two  hundred  and 
thirty ;  four  hundred  and  fifty ;  six  hundred  and  seventy. 


14 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


5.  One  hundred  and  fifty-three;  four  hundred  .and 
eighty-six;  seven  hundred  and  twenty-nine;  one  hun- 
dred and  three;  four  hundred  and  six;  seven  hundred 
and  nine. 


NUMBERS    TO    BE    READ. 


1. 

210 

320 

430 

540; 

650 

760. 

2. 

213 

546 

879 

417; 

528 

639. 

3. 

201 

435 

768 

324; 

657 

980. 

4. 

543 

876 

192 

329; 

548 

765. 

5. 

513 

846 

279 

301; 

604 

907. 

UNITS  OF  HIGHER  ORDERS. 

8.  1.  Ten  hundreds  are  one  thousand;  it  is  repre- 
sented by  1  in  the  fourth  order;  thus,  1000. 

2.  Ten  thousands  form  a  unit  of  the  fifth  order ;  thus, 
10000 ;  one  hundred  thousands,  a  unit  of  the  sixth  order ; 
thus,  100000,  etc. 

3.  Invariably,  ten  units  of  any  order  make  a  unit  of  the 
next  higher  order. 

4.  The  names  of  the  first  nine  orders  may  be  learned 
from  the  followino- 


Table  of  Orders. 
9th.     8th.     7th.     6th.     5th.     4th.      3d.      2d.      Ist 


'T3 


t 


o 


T3 
0 


O 


1^       W       H 


C 
o 


^ 

a 


o 


NOTATION.  15 


DEFINITIONS  AND  PRINCIPLES. 

9.  1.  The  first  nine  numbers  are  represented  by  the 
nine  figures, — 1,  2,  3,  4,  5,  6,  7,  8,  9. 

2.  All  other  numbers  are  represented  by  combinations 
of  two  or  more  of  the  ten  figures, — 1,  2,  3,  4,  5,  6,  7,  8, 
J,  0. 

3.  The  numbers  that  end  with  2,  4,  6,  8,  or  0  are 
called  even  numbers. 

4.  The  numbers  that  end  with  1,  3,  5,  7,  or  9  are 
called  odd  numbers. 

5.  The  value  of  a  figure  is  the  number  of  units  it  ex- 
presses. 

6.  The  value  of  a  figure  is  always  local;  that  is,  it 
depends  upon  the  place  it  occuj)ies  in  a  number. 

Rem. — The  principle  of  local  value  is  Avhat  peculiarly  distinguishes 
the  Arabic  System  of  Notation  from  all  other  systems  that  have 
existed. 

7.  The  number  a  figure  expresses  w^hen  it  stands  in 
units'  place  is  called  its  simple  value. 

8.  The  value  of  a  figure  is  increased  tenfold  by  remov- 
ing it  one  place  to  the  left. 

9.  The  value  of  a  figure  is  decreased  tenfold  by  remov- 
ing it  one  place  to  the  right. 

GROUPING  OF  ORDERS  INTO  PERIODS. 

10.  1.  For  convenience  in  writing  and  reading  num- 
bers, the  diflPerent  orders  are  grouped  into  periods  of 
three  orders  each. 

Rem. — A  number  is  pointed  off  into  periods  of  three  figures  each 
by  commas. 

2.  The  first  three  orderp — units,  tens,  hundreds — con- 
stitute the  first,  or  unit  period. 


16 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


3.  The  second  group  of  three  orders — thousands^  ten 
thousands,  hundred  thousands — constitutes  the  second,  or 
thousand  period. 

4.  The  third  group  of  three  orders  constitutes  the 
third,  or  million  period. 

5.  The  periods  from  the  first  to  the  twelfth  inclusive 
may  be  learned  from  the  following 

Table  of  Periods. 


No. 

Name. 

No. 

Name. 

First 

Unit. 

Seventh 

Quintillion. 

Second 

Thousand. 

Eighth 

Sextillion. 

Third 

Million. 

Xinth 

Septillion. 

•  Fourth 

Billion. 

Tenth 

Octillion. 

Fifth 

Trillion. 

Eleventh 

Nonillion. 

Sixth 

Quadrillion. 

Twelfth 

Decillion. 

6.  The  grouping   of  the    orders  into   periods  is  shown 
in  the  following 

Table. 


9 


o 


en 


4. 

Billion. 


o 

a?  '-^ 
^  IS 


o 

S  H  pq 


^    S  B 


3. 

Million 


o 


mII 


2.  1. 

Thousand.  Unit. 


TU 

fl 


't:! 
o 

^ 


O 


W  H  H 


3      3     G 

H  H  P 


NOTATION.  17 

7.  It  is  plain  that  each  period  is  composed  of  unitSj 
lens  J  and  hundreds  of  that  iwriod. 

To  Write  Xumhers  in  the  Arabic  System, 

11.  1.  Write  six  hundred  and  fifty -four  trillion  three 
hundred  and  twenty -one  billion  nine  hundred  and  eighty- 
seven  million  six  hundred  and  fiftj-four  thousand  three 
hundred  and  twenty-one. 

-d 

g  ^-  G  c3 


o 


o 


6   5   4,         3   2    1,         1)   8   7,         6   5   4, 


3   g   a 

-«    0=  5 

^       OJ       ^ 

WhP 

Wh^ 

^ 
G 


s  •: 


c 


p 


MhP       WhP       WhP 

Rule. — Begin  at  the  left,  and  write  each  period  as  a 
number  composed  of  himdreds,  tens,  and  units — filling  the 
vacant  orders  with   ciphers. 

Rem. — In  the  left  hand  period,  however,  when  the  hundreds  or  the 
hundreds  and  tens  are  wanting,  the  vacant  orders  are  not  filled  with 
ciphers. 

NUMBERS   TO   BE   WRITTEN. 

2.  Two  thousand;  thirty  thousand,  four  hundred  thou- 
sand. 

3.  Five  million ;  sixty  million  ;  seven  hundred  million. 

4.  Eight  billion;  ninety  billion;  one  hundred  billion. 


18  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

5.  One  thousand  two  hundred ;  two  thousand  one  hun- 
dred. 

G.  Three  thousand  four  hundred  and  fifty ;  six  thou- 
sand  seven  hundred  and  eighty-nine. 

7.  Tw^elve  thousand  three  hundred  and  forty-five. 

8.  Six  hundred  and  seventy-eight  thousand  nine  hun- 
dred and  twelve. 

9.  One  milHon  three  hundred  and  fil'ty-seven  thou- 
sand nine  hundred  and  twenty-four. 

10.  vSixty-eight  million  one  hundred  and  forty-three 
thousand  seven   hundred  and  ninety-two. 

11.  One  thousand  and  one;  one  thousand  and  ten; 
one  thousand  one  hundred. 

12.  One  thousand  one  hundred  and  one ;  one  thousand 
one  hundred  and  ten ;  one  thousand  one  hundred  and 
eleven. 

13.  Two  thousand  and  three;  four  thousand    and   fifty. 

14.  Forty -five  thousand  and  tw^enty-six. 

15.  Eighty  thousand  two  hundred  and  one. 

16.  Ninety  thousand  and  one. 

17.  Four  hundred  and  ten  thousand  two  hundred  and 
five. 

18.  One  hundred  thousand  and  ten. 

19.  Three  million  seventy  thousand  five  hundred  and 
nine. 

20.  Forty -five  million  eighty -three  thousand  and 
twenty-six. 

21.  Nine  hundred  and  nine  million  ninety  thousand. 

22.  Seven  hundred  million  ten  thousand  and  tw^o. 

23.  Forty  billion  two  hundred  thousand  and  five. 

24.  Seven  hundred  and  twenty-six  billion  fifty  million 
une  thousand  two  hundred  and  forty-three. 

25.  Eighty  billion  seven  hundred  and  three  million 
five  hundred  and  four. 


NOTATIOJ^.  19 

12.  Numeration  is  the  reading   of  numbers  when  ex- 
pressed according  to  a  system  of  notation. 

To  Read  JYutnbers  in  the  Arabic  Sr/stein. 
1.  Eead  654321987654321. 


I  § 

HP  1^  H  0 


6    5    4,         3    2    1,         9    8    7,         6    5   4,         3   2   1. 


a, 


V. 

Of 

.-§ 

^ 

fl 

1— ' 

O 

H 

HP 

OD     +J 


W^p       W^P       W^P       H^P       wIp 

Rule. — 1,  Begin  at  the  right,  and  point  off  the  number  info 
periods  of  three  figures  each. 

2.  Begin  at  the  left,  and  read  each  jjeriod  as  a  number  com- 
posed of  hundreds,  tens,  and  units,  giving  the  name  of  the 
period. 

Rem.  1. — The  left  hand  period  will  sometimes  contain  hut  one  oi 
two  figures. 

Rem.  2. — It  is  customarj^  to  omit  the  name  of  the  unit  period. 

NUMBERS  TO  BE  READ. 

2.  41582;  763491;  2519834;  375486921;  4923176358. 

3.  37584216974;  432685729145;  6253971438267. 

4.  1300;  2540;  6070;  8009;  13200;  1005. 

5.  682300;  8600050;  3040;  50004;  704208. 

6.  7085;  62001;  400009;  2102102;  9001003. 


20  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

7.  130670921;  6900702003;  23004090701;  9420163070. 

8.  570000010326049;  200103478511992485. 

9.  45763000020108000507. 
10.  800820020802008. 

THE  ROMAN  SYSTEM  OF  NOTATION. 
DEFINITIONS. 

13.  1.  To  express  numbers,  the  Roman  Notation  em- 
ploys seven  letters;  namely,  I,  V,  X,  L,  C,  D,  M. 

Rem. — The  Roman  System  of  Notation  is  so  called  because  it  was 
the  method  of  expressing  numbers  used  by  the  ancient  Romans.  It 
is  now  used  to  mark  the  chapters  of  books,  the  dial  plates  of  clocks, 

etc. 

2.  In  the  Koman  Notation,  numbers  arc  expressed  in 
four  ways. 

1st.  Each  of  the  seven  letters  expresses  a  numher,  as  fol- 
lows: I,  one;  V,  five;  X,  ten;  L,  fifty;  C,  one  hundred ; 
D,  five  hundred ;  M,  one  thousand. 

2d.  Seven  numbers  are  expressed  by  repetitions  of  the  let- 
ters 7,  JT,  and  C.  Thus,  II  represent  two;  Hi,  three; 
XX,  twenty ;  XXX,  thirty;  CC,  two  hundred;  CCC,  three 
hundred ;  CCCC,  four  hundred. 

3d.  Four  numbers  are  expressed  by  a  subtractive  combi- 
nation, as  follows:  IV,  four;  IX,  nine;  XL,  forty;  XC, 
ninety. 

4th.  All  other  numbers  are  formed  by  additive  combina- 
tions of  two  or  more  of  the  preceding  eighteen  numbers,  the 
smaller  being  always  situated  on  the  right  of  the  larger 
number. 

For  example,  YI  is  six;  XYII,  seventeen;  LXXYIII, 
seventy -eight ;  CLXXXIX,  one  hundred  and  eighty-nine  \ 
MDCCCLXXYII,  eighteen  hundred,  and  seventy -seven. 


NOTATION. 


21 


Write  in  the  Koman  J^otation, 

1.  The  numbers  from  one  to  twenty. 

2.  The  numbers  from  twenty  to  thirty. 


3.   30 

40 

50 

60; 

70 

80; 

90. 

4.   57 

29, 

61 

38; 

46 

72; 

93. 

5.  100 

101; 

106 

117; 

129 

168. 

6.  199 

246, 

309 

482; 

527 

693. 

7.  734 

859 

975 

1001; 

1010. 

8.  1048 

1119 

1285 

1326. 

9.  1492, 

1776; 

1861 

1900. 

THE  FUNDAMENTAL   RULES. 
DEFINITIONS. 

14.     1.  An  integer  is  a  whole  number. 

2.  Numbers  are  either  abstract  or  concrete. 

3.  An  abstract  number  is  a  number  simply,  as  5, 
12,  20. 

4.  A  concrete  number  is  a  number  applied  to  one 
or  more  objects;    as  1  apple,  5  pounds,  12  men. 

5.  The  name  of  the  object  of  a  concrete  number  is  its 
denomination.  Thus,  in  5  pounds,  the  denomination  is 
pounds. 

6.  Numbers  are  either  simple  or  compound. 

7.  A  simple  number  is  a  single  number,  either  abstract 
or  concrete;  as  3,  7  dollars,  1  pint. 

8.  A  compound  number  is  made  up  of  two  or  more 
concrete  numbers  of  different  denominations ;  as  3  pecks 
7  quarts  1  pint. 

9.  There  are  four  primary  operations  of  Arithmetic; 
namely.  Addition^  Subtraction,  Multiplicatio7i,  and  Division; 
— these  are  called  the  Fundamental  Rules. 


15.  1.  If  you  have  2  cents  and  find  3  cents,  how 
many  will  you  then  have?  Ans.  5  cents. 

Why?     Because  2  cents  and  3  cents  are  5  cents. 

2.  I  spent  12  cents  for  a  slate,  and  5  cents  for  a  copy- 
book :   how  many  cents  did  I  spend  ? 

Ans.  17  cents.     Why? 

3.  John  gave  6  cents  for  an  orange,  7  cents  for  pen- 
cils, and  9  cents  for  a  ball :  how  many  cents  did  all 
cost?  A71S.  22  cents.     Why? 

4.  Joseph  gave  5  cents  for  a  daily  paper,  10  cents  for 
a  weekly  paper,  25  cents  for  a  monthly  magazine,  30 
cents  for  a  book  of  poems,  and  40  cents  for  a  novel : 
how  much  did  he  spend?  110  cents. 

16.  1.  The  operation  in  these  examples  is  termed  Ad- 
dition;  hence,  Addition  is  the  process  of  uniting  two  or 
more  numbers  into  one  number. 

2.  The  number  obtained  by  addition  is  the  Sum  or 
Amount. 

3.  When  the  numbers  to  be  added  are  simple,  the 
operation  is  called  Addition  of  Simple  Numbers. 

4.  The  sign  of  Addition  (-|-),  called  j)lus,  means  more; 
when  placed  between  two  numbers,  it  shows  that  they 
are  to  be  added ;  thus,  4  +  2  means  that  4  and  2  are  to 
be  added  together. 

(22) 


ADDITION  OF  SIMPLE  NUMBERS. 


23 


5.  The  8ign  of  equality  (==:)  denotes  that  the  quantities 
between  which  it  sstands  are  equal;  thus,  the  expression 
4  +  2  =  6  means  that  the  sum  of  4  and  2  is  6  :  it  is 
read,  4  plus  2  equals  6. 

Addition  Table. 


2  +  0=  2 

3  +  0=  3 

4  +  0=  4 

5  +  0=  5 

2  +  1=  3 

3  +  1=  4 

4+1=  5 

5  +  1=  6 

2  +  2=  4 

3  +  2=  5 

4+2=  6 

5  +  2=  7 

2  +  3=  5 

3  +  3=  6 

4  +  3=  7 

5  +  3=  8 

2  +  4=  6 

3  +  4=  7 

4  +  4=  8 

5  +  4=  9 

2  +  5=  7 

3  +  5=  8 

4  +  5=  9 

5  +  5  =  10 

2  +  6=  8 

3  +  6=  9 

4  +  6  =  10 

5  +  6  =  11 

2  +  7=  9 

3  +  7  =  10 

4  +  7  =  11 

5  +  7  =  12 

2  +  8  =  10 

3  +  8  =  11 

4  +  8  =  12 

5  +  8  =  13 

2  +  9  =  11 

3  +  9  =  12 

4  +  9  =  13 

5  +  9  =  14 

6  +  0=  6 

7  +  0=  7 

8  +  0=  8 

9  +  0=  9 

6  +  1=  7 

7  +  1=  8 

8  +  1=  9 

9  +  1  =  10 

6  +  2=  8 

7  +  2=  9 

8  +  2  =  10 

9  +  2  =  11 

6  +  3=  9 

7  +  3  =  10 

8  +  3=11 

9  +  3  =  12 

6  +  4  =  10 

7  +  4  =  11 

8  +  4  =  12 

9_[_4:^13 

6  +  5  =  11 

7  +  5  =  12 

8  +  5  =  13 

9  +  5  =  14 

6  +  6  =  12 

7  +  6  =  13 

8  +  6  =  14 

9  +  6  =  15 

6  +  7  =  13 

7  +  7  =  14 

8  +  7  =  15 

9  +  7  =  16 

6  +  8  =  14 

7  +  8  =  15 

8  +  8  =  16 

9  +  8  =  17 

6  +  9  =  15 

7  +  9  =  16 

8  +  9  =  17 

9  +  9  =  18 

17.     When  the    sum    of  the  figures  in  a  eokimn    does 
not  exceed  9,  it  is  written  under  the  column  added. 


24  KAY'S  NEW  PRACTICAL  AKITHMKTIC. 


Examples. 

1.  I  own  3  tractH  of  land :  the  first  contains  240  acres ; 
the  second,  132  acres;  the  third,  25  acres:  how  many 
acres  in  all? 

SoLUTiOiV. — Since  units  oi different  orders  can  not  be  added  together, 
write  units  of  the  same  order  in  the  same  column,  so  that  the  figures 
to  be  added  may  be  in  the  most  convenient  ])ositio7i. 

Begin  at  the  right,  and  say  5  and  2  are  7  units,  2  4  0  acres, 

which  write  in  units'  place;  2  and  3  are  5,  and  4  132  acres, 

are  9  tens,  which  write  in  tens'  place;  1   and  2  are  3  2  5  acres, 

hundreds,  which  write  in  hundreds' place.  39  7  acres. 

2.  I  owe  one  man  $210,  another  S142,  and  another 
$35:  what  is  the  sum  of  my  debts?  $387. 

3.  Find  the  sum  of  4321,  1254,  3120.  8695. 

4.  Find  the  sum  of  50230,  3105,  423.  53758. 

18.  When  the  sum  of  the  figures  in  a  column  ex- 
ceeds 9,  two  or  more  figures  are  required  to  exi)re8S  it. 


Example. 
1.  Add  the  numbers  3415,  503,  1870,  and  922. 


3415 
503 


Solution. — Write  units  of  the  same  order  in  the  same 
column.     Then    say  2  and  3  are  5,  and  5  are  10  units, 
which  are  no  (  0  )  units,  written  in  the  units'  place,  and  1      18  70 
ten,  carried  to  the  tens;  1  and  2  are  3,  and  7  are  10,  and  1         922 
are  11  tens,  which  are  1  ten,  written  in  the  tens'  place,  and      6  710 
1  hundred,  carried  to  the  hundreds;  1  and  9  are  10,  and  8 
are  18,  and  5  are  23,  and  4  are  27  hundreds,  which  are  7  hundreds, 
written  in  the  hundreds'  place,  and  2  thousands,  carried  to  the  thou- 
sands; 2  and  1    are  3,  and  3  are  6  thousands,  written  in  the  thou- 
sands' place. 


4 


ADDITION  OF  SIMPLE  NUMBERS.  25 

CaiTying  the  tens  is  simply  adding  tens  to  tens,  hun- 
dreds to  hundreds,  etc.,  on  the  principle  that  only  units 
of  the  same  order  can  be  added. 

For  convenience,  the  addition  begins  at  the  right  hand 
column,  with  the  units  of  the  lowest  order,  so  that,  if 
the  sum  of  the  figures  in  any  column  exceeds  9,  the  tens 
can  be  carried  to  the  sum  of  the  next  higher  order. 

Rem. — To  illustnitc;  the  greater  convenience  of  adding  the  units' 
column  first,  take  the  above  example. 

Solution. — Commencing  the  addition  with  the  thou-  3415 
sands'  column,  the  sum  is  4;  next  adding  the  hundreds,  ^^^ 
the  sum  is  26  hundreds,  which  equal  2  thousands  and  6  f.<^o 
hundreds;  next  adding  the  tens,  the  sum  is  10  tens,  equal 
to  1  hundred;  and  finally  adding  the  units,  the  sum  is  10 
units,  equal   to   1    ten.     As  these  sums   have    also   to   be         10 

added,  this   much  extra  work  must  be  done  in  order  to  __1_9 

complete  the  solution.  6710 

19.  Rule. — 1.  Write  the  mnnhers  to  he  added,  so  that 
figures  of  the  same  order  may  stand  in  the  same  coluinn. 

2.  Begin  at  the  right  hand,  and  add  each  column  sepa- 
rately. Place  the  units  obtained  by  adding  each  cohwm 
under  it,  and  carry  the  tens  to  the  next  higher  order.  Write 
down  the  entire  sum  of  the  last  column. 

Proof. — Add  the  columns  downward,  commencing  with 
the  column  of  units. 

1.  Find  the  sum  of  3745,  2831,  5983,  and  7665. 

In    adding   long   cokimns   of    figures,  it   is   necessary        3  745 

to  retain  the  numbers  carried.      This  may  be   done   by        2831 

5  9  83 
placing    them    in   smaller    figures    under    their   proper        7«ak 

columns,  as  3,  2,  1,  in  the  margin.  9  0  9  2  4 

321 


26  liAY'tS  NEW  PRACTICAL  AKITHMETIC. 


Examples. 

(2)  (3)  (4)  (5)  (6)  (7) 

184  204  103  495  384  1065 

216  302  405  207  438  6317 

135  401  764  .^-85  348  5183 

320  311  573  825  843  7102 

413  109  127  403  483  3251 

101  43  205  325  834  6044 


1369 

1370 

2177 

2440     3330 

28962 

(8) 

(9) 

(10) 

(H) 

(12) 

3725 

5943 

82703 

987462 

6840325 

5834 

6427 

102 

478345 

7314268 

4261 

8204 

6005 

610628 

3751954 

7203 

7336 

759 

423158 

6287539 

13.  11  +  22  +  33  -(-  44  +  55  =  how  man}^  ?  165. 

14.  23  +  41  -I  74  +  83  +  16  =  how  many  ?  237. 

15.  45 +  19 +  32 +  74 +  55==  how  many?  225. 

16.  51  +  48  +  76  +  85  +    4  =  how  many  ?  264. 

17.  263  +  104  +  321  +  155  =  how  many  ?  843. 

18.  94753  +  2847  +  93688  +  9386  +  258  +  3456  are 
how  many  ?  204388. 

19.  January  has  31  days ;  February,  28 ;  March,  31 ; 
April,  30 ;  and  May,  31 :  how  many  days  are  there  in 
these  five  months?  151. 

20.  June  has  30  days;  July,  31;  August,  31;  September, 
30;  October,  3l :  how  many  days  in  all?  153. 

21.  The  first  5  months  have  151  days,  the  next  5  have 
153  days,  November  has  30,  and  December,  31 :  how 
many  days  in  the  whole  year?  365. 


ADDITION  OF  SIMPLE  NUMBEES.  27 

22.  I  bought  4  pieces  of  muslin :  the  first  contained  50 
yards,  the  second,  65,  the  third,  42,  the  fourth,  89 :  how 
many  yards  in  all?  246  yd. 

23.  I  owe  one  man  S245,  another  $325,  a  third  $187, 
a  fourth  $96 :  how  much  do  I  owe  ?  $853. 

24.  General  Washington  was  born  A.  D.  1732,  and 
lived  67  years:  in  what  year  did  he  die?  1799. 

25.  Alfred  the  Great  died  A.  D.  901 ;  thence  to  the 
signing  of  Magna  Charta  was  314  years ;  thence  to  the 
American  Eevolution,  560  years:  in  what  year  did  the 
American  Eevolution  begin?  1775. 

26.  A  has  4  flocks  of  sheep ;  in  the  first  are  65  sheep 
and  43  lambs ;  in  the  second,  187  sheep  and  105  lambs ; 
in  the  third,  370  sheep  and  243  lambs ;  in  the  fourth, 
416  sheep  and  95  lambs:  how  many  sheep  and  lambs 
has  he?  1038  sheep,  and  486  lambs. 

27.  A  man  bought  30  barrels  of  pork  for  $285,  18 
barrels  for  $144,  23  barrels  for  $235,  and  34  barrels  for 
$408 :  how  many  barrels  did  he  buy,  and  how  many 
dollars  did  he  pay?  105  bbl.,  and  $1072. 

28.  The  first  of  four  numbers  is  287 ;  "^^^he  second,  596 ; 
the  third,  841 ;  and  the  fourth,  as  much  as  the  first  three : 
what  is  their  sum?  3448. 

29.  The  Pyramids  of  Egypt  were  built  1700  years 
before  the  founding  of  Carthage ;  Carthage  was  founded 
47  years  before  and  was  destroyed  607  years  after  the 
founding  of  Eome,  or  146  years  before  the  Christian  era. 
How  many  years  before  Christ  were  the  Pyramids 
built?  2500. 

30.  Add  three  thousand  and  five;  fort^^-two  thousand 
six  hundred  and  twenty-seven  ;  105 ;  three  hundred  and 
seven  thousand  and  four;  80079;  three  hundred  and 
twenty  thousand  six  hundred.  753420. 

31.  Add  275432 ;    four  hundred  and  two  thousand  and 


28  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

thirty ;  three  hundred  thouBand  and    five ;    872026 ;    four 
million  two  thousand  three  hundred  and  forty -seven. 

5851840. 

32.  Add  eight  hundred  and  eighty  million  eight  hun- 
dred and  eighty-nine ;  2002002 ;  seventy-seven  million 
four  hundred  and  thirty-six  thousand;  two  hundred  and 
six  million  five  thousand  two  hundred  and  seven  ;  49003; 
nine  hundred  and  ninety  million  nineteen  thousand  nine 
hundred  and  nineteen.  2155513020. 

33.  North  America  has  an  area  of  8955752  square 
miles;  South  America,  6917246  square  miles;  and  the 
West  Indies,  94523  square  miles:  what  is  the  area 
of  the   entire   continent?  15967521  sq.  mi. 

34.  A  man  pays  $600  for  a  lot,  $1325  for  building 
materials,  $30  for  digging  the  cellar,  $120  for  stone- 
work, $250  for  brick-work,  $140  for  carpenter- work, 
$120  for  plastering,  and  $115  for  painting:  how  much 
did  his  house  and  lot  cost  him?  $2700. 

35.  A  man  bequeaths  $7850  to  his  wife,  $3275  to  each 
of  his  two  sons,  and  $2650  to  each  of  his  three  daughters : 
what  is  the  amount  of  his  bequest?  $22350. 

36.  A  merchant  spent  $8785  for  dress  goods,  and  $12789 
for  sheetings.  He  sold  the  dress  goods  at  a  profit  of  $878, 
and  the  sheetings  at  a  profit  of  $1250:  for  how  much  did 
he  sell  the  whole?  $23702. 

37.  A  merchant  began  business  with  $7000  cash,  goods 
w^orth  $12875,  bank  stock  worth  $5600,  and  other  stocks 
worth  $4785.  In  one  year  he  gained  $3500 :  what  w^as 
he  worth  at  its  close?  $33760. 

38.  A  house  has  two  parlors,  each  requiring  30  yards 
of  carpet;  four  bed-rooms,  each  requiring  25  yards;  a 
dining-room  and  sitting-room,  each  requiring  20  yards: 
how  many  yards  are  required  to  carpet  the  entire 
house?  200  yd. 


ADDITION  OF  SIMPLE  NUMBERS. 


29 


20.  An  excellent  practice,  in  order  to  secure  readiness 
and  accuracy,  is  to  add  two  columns  at  once.  The  fol- 
lowing example  illustrates  the  method : 


(!)■ 


Beginning  with  47,  add  the  3  tens  above,  which  equal 
77  ;  then  the  4  units,  making  81 ;  then  the  6  tens  above, 
141;  and  the  5  units,  146;  then  the  7  tens  above,  216; 
and  the  9  units,  225^  then  the  9  tens  above,  315,  and 
finally  the  2  units,  317.  Put  down  the  17,  and  carry  the 
3  hundreds  to  the  hundreds'  column.  Then  93  and  3  to 
carry  are  96,  and  60  are  156,  and  2  are  158,  and  40  are  198, 
and  8  are  206,  and  60  are  266,  and  7  are  273,  and  70  are  343,  and 
are  351,  wnk'h  write  in  its  proper  place. 


7892 
6  7  79 
4865 
6234 
9347 
35117 


Examples. 

(2) 

(3) 

w 

(5) 

(H) 

3686 

9898 

4356 

893742 

234567 

4724 

8989 

6342 

743698 

765432 

6583 

4545 

7989 

437821 

987654 

5798 

5454 

4878 

643567 

456789 

6953 

6363 

6749 

892742 

778899 

27744 


35249 


30314 


3611570 


3223341 


(7) 


(8) 


(9) 


(10) 


5493275 

4819 

18356 

849627 

6182463 

9263 

49276 

532472 

9538719 

2752 

94678 

293784 

2645834 

8375 

36525 

468135 

8256386 

6498 

42983 

926547 

32116677 

31707 

241818 

3070565 

30 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


(11) 

(12) 

(13) 

(14) 

(15) 

7421 

6873 

4729 

237285 

884261 

6322 

2196 

6234 

64371 

724353 

798 

583 

5781 

2143 

416213 

4352 

79 

3143 

842 

598624 

547 

684 

7182 

55 

784344 

674 

4348 

6989 

789 

627517 

2315 

7896 

7222 

4621 

843641 

7218 

233 

6643 

15115 

47821 

1847 

594 

7859 

647890 

52348 

5721 

6483 

6742 

77442 

2932 

6848 

7542 

8982 

84931 

4751 

4722 

3967 

3451 

894623 

896 

5976 

29 

8692 

446217 

722 

6843 

478 

7341 

134162 

823344 

1234 

1717 

6822 

192317 

874132 

62833 


43702 


97812 


2802803 


6685899 


21.  1,  If  you  have  9  apples,  and  give  4  away,  how 
many  will  you  have  left?  Ans.  5  apples. 

Why?     Because  4  apples  from  9  apples  are  5  apj^les. 

2.  Frank  had  15  cents ;  after  spending  7,  how  many 
were  left?  Aiis.  8  cents.     Why? 

3    If  you  take  8  from  13,  how  many  are  left?     Ans.  5. 

4.  If  I  have  25  cents,  and  spend  10  of  them  for  a  lead- 
l^encil,  how  much  will  I  have  left  ?  Ans.  15  cents. 

5-  Twelve  from  twenty  leaves  how  many?  A71S.  8. 

22.  1.  The  operation  in  the  preceding  examples  is 
termed  Subtraction ;  hence.  Subtraction  is  the  process  of 
finding  the  difference  between  two  numbers. 

2.  The  larger  number  is  called  the  Minuend;  the  less, 
the  Subtrahend ;  and  the  number  left  after  subtraction, 
the  .Difference  or  Remainder. 

3.  When  the  given  numbers  are  simple,  the  operation 
is  called  Subtraction  of  Simple  Numbers. 

23.  The  sign  of  Subtraction  ( — ;  is  called  minus, 
meaning  less;  when  placed  between  two  numbers,  it  de- 
notes that  the  number  on  the  right  is  to  be  taken  from 
the  one  on  the  left;  thus,  8  —  5  =  3  means  that  5  is 
to  be  taken  from  8,  and  is  read,  8  minus  5  equcds  3. 

(31) 


32 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Subtraction  Table. 


2  —  2  =  0 

3  —  3  =  0 

4  —  4  =  0 

5-5  =  0 

3-2  =  1 

4  —  3  =  1 

5-4  =  1 

6-5  =  1 

4-2  =  2 

5  —  3  =  2 

6  —  4  =  2 

7  —  5  =  2 

5  —  2  =  3 

6  —  3  =  3 

7  —  4  =  3 

8  —  5  =  3 

6  —  2  =  4 

7-3  =  4 

8-4  =  4 

9-5  =  4 

7  —  2  =  5 

8  —  3  =  5 

9  —  4  =  5 

10  —  5  =  5 

8  —  2  =  6 

9  —  3  =  6 

10  —  4  =  6 

11  —  5  =  6 

9  —  2  =  7 

10-3  =  7 

11  —  4  =  7 

12  —  5  =  7 

10  —  2  =  8 

11  —  3  =  8 

12  —  4  =  8 

13  —  5  =  8 

11  —  2  =  9 

12  —  3  =  9 

13  —  4  =  9 

14  —  5  =  9 

6  —  6  =  0 

7  —  7  =  0 

8  —  8  =  0 

9  —  9  =  0 

7--6  =  l 

8  —  7  =  1 

9  —  8  =  1 

10  —  9  =  1 

8  —  6  =  2 

9-7  =  2 

10  —  8  =  2 

11  —  9  =  2 

9  —  6  =  3 

10  —  7  =  3 

11  —  8  =  3 

12  —  9  =  3 

10  —  6  =  4 

11  —  7  =  4 

12  —  8  =  4 

13  —  9  =  4 

11  —  6  =  5 

12  —  7  =  5 

13  —  8  =  5 

14  —  9  =  5 

12  —  6  =  6 

13  —  7  =  6 

14  —  8  =  6 

15-9  =  6 

13  —  6  =  7 

14  -  7  =  7 

15  — 8-=  7 

16  —  9  =  7 

14  —  6  =  8 

15  —  7  =  8 

16  —  8  =  8 

17-9  =  8 

15  —  6  =  9 

16  —  7  =  9 

17  —  8  =  9 

18  —  9  =  9 

24.   When  each  figure  of  the  subtrahend  is  not  greater 
than  the  corresponding  figure  of  the  minuend. 


Examples. 

1.  A  man  having  $135,  spent  $112:  how  much  had  he 
left? 


SUBTRACTION  OF  SIMPLE  NUMBERS.  33 

Solution. — Since  the  difference  between  units  of  the  same  order 
only  can  be  found,  write  units  of  the  same  order  in  the  same 
column,  so  that  the  figures  between  which  the  subtraction  is  to  be 
made  may  be  in  the  most  convenient  position. 

Begin  at  the  right,  and  say  2  from  5  leaves 
3,  which  put   in  units'  place;  I  from  3  leaves       ^^^  minuend. 
2,   which  put  in  tens'  place;  1   from  1  leaves      il^  subtrahend. 
0,  and,  there  being  no  figures  on  the  left  of         ^^  remamder. 
this,  the  place  is  vacant. 


2.  A  farmer  having    245  sheep,  sold    123:  how    many 
sheep  had  he  left?  122. 

3.  A  man   bought   a   farm   for   $751,  and  sold    it   for 
$875:  how  much  did  he  gain?  $124. 

What  is  the  difference 

4.  Between  734  and  531?  203. 

5.  Between  8752  and  3421?  5331. 

6.  Between  79484  and  25163?  54321. 

7.  Between  49528  and  16415?  *      33113. 

25.    When    the   lower   figure   in   any  order   is   greater 
than   the   upper,  a   difficulty  arises,  which   we   will  now 


explain. 


Examples. 


1.  James  had    13   cents;    after  spending  5,  how  manv 
cents  had  he  left? 

1  3 
5  can  not  be  subtracted  from  3,  but  it  can  be  from  13;  ^ 

6  from  13  leaves  8.  "qT 


2.  From  73  subtract  45. 

Prac.  3. 


34  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

Solution. — 5  units  can  not  be  taken  from  3  units. 
Take  1  (ten)  from  the  7  (tens),  and  add  this  1  (ten)  or  7  3 

10  units  to  the  3  units,  which  makes  13  units;  then,  4  5 

subtract  the  5  units,  and  there  will  remain  8  units,  to  be  2  8 

put  in  units'  place.     Since  1  ten  is  taken  from  the  7 
tens,  there  remain  but  6  tens.     Subtract  4  tens  from  6  tens  and  pui 
the  remainder,  2  tens,  in  tens'  place.     The  difference  is  28. 

Rem.  1. — Instead  of  actually  taking  1   ten   from  the  7  tens,  and 
adding  it  to  the  3  units,  the  operation  is  perfi.rmed  mentally;  thus, 
6  from  13  leaves  8,  and  4  from  6  leaves  2. 

Rem.  2. — In  such  cases,  the  value  of  the  upper  number  is  not 
changed,  since  the  1  ten  which  is  taken  from  the  order  of  tens  is 
added  to  the  number  in  the  order  of  units. 

Rem.  3. — Taking  a  unit  of  a  higher  order  and  adding  it  to  the 
units  of  the  next  lower,  so  that  the  figure  beneath  may  be  subtracted 
from  the  sum,  is  called  borrowing  ten. 

Rem.  4. — After  increasing  the  units  by  10,  instead  of  considering 
the  next  figure  of  the  up}>er  number  as  diminished  by  1,  the  result 
will  be  the  same,  if  the  next  figure  of  the  lower  number  be  increased 
by  1;  thus,  in  the  previous  example,  instead  of  diminishing  the  7 
tens  by  1,  add  1  to  the  4  tens,  which  makes  5;  thus,  5  from  13  leaves 
8,  and  5  from  7  leaves  2. 

Rem.  5. — This  process  depends  upon  the  fact  that  having  borrowed 
1  from  the  7  tens,  we  have  to  subtract  from  it  lx)th  1  ten  and  4  tens, 
or  their  sum,  5  tens. 

3.  Find  the  difference  between  805  and  637. 

Solution — 1st  Method. — Writing  the   less   number  8  05 

under  the  greater,  with  units  of  the  same  order  in  the  6  3  7 

same  column,  it  is  required  to  subtract  the  7  units  from  16  8 

5  units. 

The  five  can  not  be  increased  by  borrowing  from  the  next  figure, 
because  it  is  0;  therefore,  borrow  1  hundred  from  the  8  hundreds, 
which  leaves  7  hundreds  in  hundreds'  place;  this  1  hundred  makes 
10  tens;  then,  borrowing  1  ten  from  the  10  tens,  and  adding  it  to 
the  5  units,  9  tens  will  be  in  the  tens'  place,  and  15  units  in  the 
units'  place. 


SUBTRACTION  OP  SIMrLE  NUMBERS. 


35 


Subtracting  7  from  15,  8  units  are  left,  to  be  written  in  units* 
place;  next,  subtracting  3  tens  from  9  tens,  there  are  left  6  tens,  to 
be  written  in  tens'  place;  lastly,  subtracting  6  hundreds  from  7  hun- 
dreds, there  remains  1  hundred,  to  be  written  in  hundreds'  place. 

2d  Method. — If  the  5  units  be  increased  by  10,  say  7  from 
15  leaves  8;  then,  increasing  the  3  by  1,  say  4  from  0  can  not  be 
taken,  but  4  from  10  leaves  6;  then,  increasing  6  by  1,  sa^-  7  from  8 
leaves  1. 

Rem.  1. — The  second  method  is  generally  used;  it  is  more  con- 
venient, and  less  liable  to  error,  especially  when  the  upper  number 
contains  ciphers. 

Rem.  2. —Begin  at  the  right  to  subtract,  so  that  if  any  lower 
figure  is  greater  than  the  upper,  1  may  be  borrowed  from  a  higher 
order. 

Rem.  3. — If  the  difference  of  two  numbers  be  added  to  the  less 
number,  the  sum  will  be  equal  to  the  greater.  Thus,  if  5  subtracted 
from  8  leave  3,  then  3  added  to  5  will  equal  8. 

26,  Rule. — 1.  Write  the  less  number  under  the  greater, 
placing  figures  of  the  same  order  in  the  same  column. 

2.  Beginning  at  the  right  hand,  subtract  each  figure  from 
the  one  directly  over  it,  and  write  the  remainder  beneath. 

3.  If  the  loiver  figure  exceeds  the  upper,  add  ten  to  the 
upper  figure,  subtract  the  lower  from  it,  and  carry  one  to  the 
next  lower  figure,  or  take  one  from  the  next  upper  figure. 

Proof. — Add  the  remainder  to  the  subtrahend ;  if  the 
sum  is  equal  to  the  minuend,  the  work  is  correct. 


Minuends, 
Subtrahends, 

(1) 
7640 
1234 

6406 

7640 

Examples. 

(2) 

860012 
430021 

(3) 
4500120 
2910221 

(4) 

3860000 
120901 

Remainders, 

429991 

1589899 

3739099 

Proof, 

860012 

4500120 

3860000 

36  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

5.  Take  1234567  from  4444444.  8209877. 

6.  Take  15161718  from  91516171.  76354453. 

7.  Take  34992884  from  63046571.  28053687. 

8.  153425178  —  53845248==?  99579930. 

9.  100000000  —  10001001==?  89998999. 

10.  Take  17  cents  from  63  cents.  46  cents. 

11.  A  carriage  cost  $137,  and  a  horse  S65 :  how  much 
more  than  the  horse  did  the  carriage  cost?  $72. 

12.  A  tree  75  feet  high  was  broken ;  the  part  that 
fell  was  37  feet  long:  how  high  was  the  stump?      38  ft. 

13.  America  was  discovered  by  Columbus  in  1492 : 
how  many  years  had  elapsed  in  1837?  345. 

14.  I  deposited  in  the  bank  $1840,  and  drew  out  $475 : 
how  many  dollars  had  1  left?  $1365. 

15.  A  man  has  property  worth  $10104,  and  owes  debts 
to  the  amount  of  $7426 :  when  his  debts  are  paid,  how 
much  will  be  left?  $2678. 

16.  A  man  having  $100000,  gave  away  $11 :  how  many 
had  he  left?  $99989. 

17.  Subtract  19019  from  20010.  991. 

18.  Eequired  the  excess  of  nine  hundred  and  twelve 
thousand  and  ten,  above  50082.  861928. 

19.  Take  4004  from  four  million.  3995996. 

20.  Subtract  1009006  from  two  million  twenty  thou- 
sand nine  hundred  and  thirty.  1011924. 

21.  Subtract  four  hundred  and  five  thousand  and 
twenty-two  from  2000687.  1595665. 

22.  What  is  the  difference  between  thirteen  million  two 
hundred  and  one  and  17102102?  4101901. 

23.  A  man  invested  in  business  $30,000;  at  the  end  of 
the  first  year  he  found  that  all  his  assets  amounted  to 
only  $26,967;  how  much  had  he  lost?  $3,033. 

24.  Take  9238715  from  18126402.  8887687. 

25.  Take  9909090009  from  19900900900.       9991810891. 


ADDITION   AND   SUBTRAQTION.  37 

Examples  in  Addition  and  Subtraction. 

1.  275  +  381  +  625—1098==?  183. 

2.  6723  —  479  —  347—   228  =  ?  5669. 

3.  In  January,  1876,  a  merchant  bought  goods  to  the 
amount  of  $2675;  in  February,  S4375;  and  in  March, 
$1897  ;  after  making  one  payment  of  $3000,  and  another 
of  $4947,  how  much  did  he  still  owe?  $1000. 

4.  I  owe  three  notes,  whose  sum  is  $1300 — one  note 
being  for  $250,  and  another  for  $650:  what  is  the  amount 
of  the  third  note?  $400. 

5.  Mr.  Jones  deposited  $450  in  bank  on  Monday;  on 
Tuesday,  $725;  oji  Wednesday,  $1235;  on  Thursday, 
$4675;  and  on  Friday,  $1727.  On  Saturday  morning  he 
drew  out  $5935,  and  Saturday  afternoon,  ^877 :  how 
much  money  had  he  left  in  bank?  $2000. 

6.  At  the  end  of  one  year  I  found  I  had  spent  $2300. 
Of  this  amount,  $350  were  paid  for  board,  $125  for  cloth- 
ing, $375  for  books,  $150  for  incidentals,  and  the  remain- 
der for  two  acres  of  ground :  how  much  did  the  two 
acres  cost?  $1300. 

7.  A  speculator  bought  three  houses.  For  the  first  he 
gave  $4875 ;  for  the  second,  $2250  more  than  for  the  first ; 
and  for  the  third  he  gave  $3725.  He  afterward  sold  them 
all  for  $20838:  how  much  did  he  gain?  $5113. 

8.  A  man  owns  j^roperty  valued  at  $49570,  of  which 
$16785  are  in  personal  property,  and  $24937  in  real 
estate ;  the  remainder  was  deposited  in  bank :  how  much 
has  he  in  bank?  $7848. 

9.  A  merchant  bought  a  bill  of  goods  for  $7895,  and 
paid  $175  for  freight,  and  $3  for  dray  age.  He  sold  the 
goods  for  $10093:  how  much  did  he  gain?  $2020. 

10.  A  farmer  invested  $10000,  as  follows:  in  land, 
$5750;  in  horses,  $925;    in  cattle,  $1575;  in  hogs,  $675; 


38  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

and  the  remainder  in  im^jlements  and  tools :    how  much 
did  he  invest  in  implements  and  tools?  81075. 

11.  A  speculator  on  Monday  gained  $4625 ;  on  Tues- 
day, $3785 ;  on  Wednesday  he  lost  $6955 ;  on  Thursday 
he  lost  $895;  on  Friday  he  gained  $985,  and  on  Satur- 
day he  lost  $1375:  how  much  did  he  gain  during  the 
entire  week?  $170. 

12.  The  following  18  Mr.  Brown's  private  account  for 
two  weeks:  First  week,  received  $50  for  salary,  and 
spent  $25  for  clothing,  $7  for  board,  $2  for  washing, 
and  $5  for  sundries.  Second  week,  received  $50  for 
salary,  loaned  $35  to  Tom  Jones,  paid  $7  for  board,  $2 
for  washing,  and  $8  for  sundries.  How  much  did  Mr. 
Brown  have  at  the  end  of  the  two  weeks?  $9. 


MULTIPLICATION  OP  SIMPLE  NUMBERS. 


43 


EXAiMPLES. 

(J) 

(8) 

(9) 

(10) 

Multiplicand, 

5142 

4184 

3172 

41834 

Multiplier, 

5 

2 

6 

5 

7 

Product, 

25710 

15104 

15860 

292838 

11.  Multiply 

49  by 

3. 

147. 

12.  Multiply 

57  by 

4. 

228. 

13.  Multiply 

128  by 

5. 

640. 

14.  Multiply 

367  by 

6. 

2202. 

15.  Multiply 

1427  by 

7. 

9989. 

16.  Multiply 

19645  by 

8. 

157160. 

17.  Multiply 

44386  by 

9. 

399474. 

18.  Multiply 

708324  by 

7. 

4958268. 

19.  Multiply 

96432  by 

10. 

964320. 

20.  Multiply 

46782  by 

11. 

514602. 

21.  Multiply 

86458  by 

12. 

1037496. 

When  the  Multiplier  Exceeds  12. 


22.  What  is  the  product  of  43  X  25  ? 

Analysis. — Since  25  is  equal  to  2  tens  and  4  3 

5   units — that   is,  20 -f  5, — multiply  by  5   and  2  5 

write  the  product,  215;  then  multiply  by  the  21  5  =  4  8  X  ^ 
2  tens,  and  set  the  product,  8  hundreds  and  6  8G  =43X20 
tens,  under  the  2  hundreds  and  1  ten.  1075  =  43X25 

Multiplying   by  5   units   gives   5   times   43, 
and  multiplying  by  2  tens  gives  20  times  43;    add  them,  because  5 
times  43  and  20  times  43  equal  25  times  43. 

Hence,  multiply  by  the  units'  figure  of  the  multiplier,  and  write 
the  product  so  that  the  right-hand  figure  will  fall  in  units'  place; 
then  multiply  by  the  tens'  figure,  and  write  the  right-hand  figure 
of  the  product  in  the  tens'  place. 


44  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

Therefore,  in  multiplying  by  a  figure  of  any  order,  write  the  last 
figure  of  the  product  in  the  same  order  as  the  multiplier. 

Note. — The  products  of  the  multiplicand  by  the  separate  figures 
of  the  multiplier  are  called  pariial  producU. 

General  Rule. — 1.  Write  the  multiplier  under  the  miilti' 

plicand,  placing  figures  of  the  same  order  in  a  column. 

2.  Multiply  the  midtiplicand  by  each  figure  of  the  multi- 
plier in  succession,  beginning  with  units,  always  setting  the 
right  hand  figure  of  each  product  under  that  figure  of  the 
multiplier  which  produces  it. 

3.  Add  the  partial  products  together :  their  sum  will  be  the 
product  sought. 

Proof. — Multiply  the  multiplier  by  the  multiplicand: 
the  product  thus  obtained  should  be  the  same  as  the 
first  product. 

23.  Multiply  2345  by  123. 

SOLUTION.  PROOr. 

123  multiplier. 

2  345  multiplicand.  234  6  multiplicand. 

12  3  multiplier.  615  =  1  2  3  X          5 

7035=^2345X       3  492     =123X       40 

469       ^2345X    20  869       =123X    300 

2345__  =2  345X1 00  246         =123X2000 

288435  =  2345X123  288435  =  123X2345 

24.  Multiply  327  by  203. 

Remark. — "When   there  is  a  cipher  in  the  multiplier,         82  7 
leave  it,  and  multiply  by  the  other  figures,  being  careful  20  3 

to  place  the  right-hand   figure  of  each   partial   product  981 

under  the  multiplying  figure.  6  5  4 

66381 


MULTIPLICATION  OF  SIMPLE  NUMBERS. 


45 


Examples. 


25.  235 

26.  34() 

27.  425 

28.  518 

29.  279 

30.  869 

31.  294 

32.  429 

33.  485 


X13  = 

X19  = 
X29  = 
X34. 
X37  = 
X49. 
X57. 
X62: 
X76: 


:  3055. 
:  6574. 
:  12325. 
:  17612. 
:  10323. 
:  42581. 
:  16758. 
=  26598. 
=  36860. 


34.  624 

35.  976 

36.  342 

37.  376 

38.  476 

39.  2187 

40.  3489 

41.  1646 

42.  8432 


X  85: 
X  97  = 
X364: 
X526  = 
X536. 
X215: 
X276  = 
X365: 
X635: 


43.  Multiply  6874  by    829. 

44.  Multiply  2873  by  1823. 

45.  Multiply  4786  by  3497. 

46.  Multiply  87603  by  9865. 

47.  Multiply  83457  by  6835. 

48.  Multiply  31624  by  7138. 


:  53040. 
94672. 
:  124488. 
:  197776. 
:  255136. 
:  470205. 
:  962964. 
:  600790. 
:  5354320. 

5698546. 

5237479. 

16736642. 

864203595. 

570428595. 

225732112. 


49.  What  will  126  barrels  of  flour  cost,  at  $6  a  bar- 
rel? $756. 

50.  What  will  823  barrels  of  pork  cost,  at  $12  a  bar- 
rel? $9876. 

51.  What  will  675  pounds  of  cheese  cost,  at  13  cents 
a  pound?  8775  cents. 

52.  What  will  496  bushels  of  oats  cost,  at  24  cents  a 
bushel?  11904  cents. 

53.  If  a  man  travel  28  miles  a  day,  how  many  miles 
will  he  travel  in  152  days?  4256  miles. 

54.  There  are  1760  yards  in  one  mile :  how  many 
yards  are  there  in  209  miles?  367840  yards. 

55.  There  are  24  hours  in  a  day,  and  365  days  in  a 
year:  if  a  ship  sail  8  miles  an  hour,  how  far  will  she 
sail  in  a  year?  70080  miles. 


46  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

56.  Multiply  two  thousand  and  twenty-nine  by  one 
thousand  and  seven.  2048203. 

57.  Multiply  eighty  thousand  four  hundred  and  one 
by  sixty  thousand  and  seven.  4824622807. 

58.  Multiply  one  hundred  and  one  thousand  and  thirty- 
two  by  20001.  2020741032. 

59.  A  grocer  bought  2  barrels  of  sugar,  each  weighing 
215  pounds,  for  8  cents  a  pound :  how  much  did  he  pay 
for  the  sugar?  3440  cents. 

60.  A  grocer  bought  a  barrel  of  molasses,  containing 
36  gallons,  for  45  cents  a  gallon ;  and  •  sold  it  for  55 
cents  a  gallon:  how  much  did  he  gain?  360  cents. 

61.  A  commission  merchant  sold  2650  bushels  of  wheat 
for  a  farmer,  at  95  cents  a  bushel,  and  charged  him  2 
cents  a  bushel  for  selling:  how  much  money  Was  duo 
the  farmer?  246450  cents. 

62.  A  farmer  bought  6  horses  of  one  man  for  75  dol- 
lars each,  and  5  horses  of  another  for  125  dollars  each, 
and  sold  them  all  for  150  dollars  each :  how  many  dol- 
lars did  he  gain?  $575. 

63.  A  merchant  bought  one  box  of  goods  for  250 
dollars,  two  more  for  325  dollars  each,  and  three  more 
for  175  dollars  each;  he  sold  them  all  so  as  to  gain 
356  dollars:  for  how  much  did  he  sell  them?  $1781. 

64.  A  farmer  bought  24  sheep,  at  5  dollars  a  head; 
36  hogs,  at  14  dollars  a  head;  and  9  cows,  at  45  dollars 
a  head :  when  he  sold  them  all,  he  lost  275  dollars :  for 
how  much  did  he  sell  them?  $754. 

65.  To  75  X  37  add  85  X  54,  and  subtract  5284.       2081. 

66.  To  69  X  53  add  48  X  27,  and  subtract  4279.         674. 

67.  I  bought  50  bags  of  coffee,  averaging  63  poupds 
in  a  bag,  paying  34  cents  a  pound :  how  much  did 
it   cost?  10719©   cents. 


MULTIPLICATION  OF  SIMPLE  NUMBERS.  47 

CONTRACTIONS   IN   MULTIPLICATION. 
CASE   I. 

32.    When  the  multiplier  can  be  separated  into  factors. 

1.  What  will  15  oranges  cost,  at  8  cents  each? 

Analysis.  —  Since     15     is     3  Cost  of    1  orange,        8  ct. 
times   5,   15   oranges  will   cost   3  5 

times   as   much   as   5   oranges.  Cost  of    5  oranges,     4  0  ct. 

Therefore,  instead  of  multiply-  3 

ing  8  by  15,  first  find  the  cost  of  Cost  of  15  oranges,  120  ct. 

6  oranges,  by  multiplying  8  cents 

by  5;    then   take   3   times  that  product  for  the  cost  of  15  oranges. 

Rule. — 1.  Separate  the  multiplier  into  tico  or  more  factors. 

2.  Multiply  the  multiplicand  by  one  of  the  factors,  and 
this  product  by  another  factor,  till  every  factor  is  used;  the 
last  product  will  be  the  one  required. 

Kem. — Do  not  confound  the  factors  of  a  number  with  the  ;?rtrts 
into  which  it  may  be  separated.  Thus,  the  factors  of  15  are  5  and  3, 
while  the  parts  into  which  15  may  be  separated  are  any  numbers 
whose  sum  equals  15:    as,  7  and  8;    or,  2,  9,  and  4. 

Examples. 

,     2.  What  will  24  acres  of  land  cost,  at  $124  an  acre? 

$2976. 

3.  How  far  will  a  ship  sail  in  56  weeks^  at  the  rate 
of  1512  miles  per  week?  84672  miles. 

4.  How  many  pounds  of  iron  are  there  in  54  loads, 
each  weighing  2873  pounds?  155142  pounds. 

5.  Multiply  2874  by  72.  206928. 

6.  Multiply  8074  by  108.  871992. 


48 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


CASE   IT. 

33.  When  the  multiplier  is  1  with  ciphers  annexed; 
as  10,  100,  1000,  etc. 

1.  Placing  one  cipher  on  the  right  of  a  number  (8, 
3)  changes  the  units  into  tens,  the  tens  into  hundreds, 
and  so  on,  and,  therefore,  multiplies  the  number  by  ten; 
thus,  annex   one   cipher  to    25,  and  it  becomes  250. 

2.  Annexing  two  ciphers  changes  units  into  hundreds, 
tens  into  thousands,  etc.,  and  multiplies  the  number  by  one 
hundred;  thus,  annex  two  ciphers  to  25,  and  it  becomes 
2500. 

Rule. — Annex  as  many  ciphers  to  the  multiplicand,  as 
there  are  ciphers  in  the  7nultiplier,  and  the  nwnber  thus 
formed  will  be  the  product  required. 


1.  Multiply      245  by  100. 

2.  Multiply      138  by  1000. 

3.  Multiply      428  by  10000. 

4.  Multiply      872  by  100000. 

5.  Multiply    9642  by  1000000. 

6.  Multiply  10045  by  1000000. 


24500. 

138000. 

4280000. 

87200000. 

9642000000. 

10045000000. 


^CASE  III. 


34.     When   there  are  ciphers  at   the  right  of  one  or 
both  of  the  factors. 

1.  Find  the  product  of  625  by  500. 


Analysis. — The    multiplier    may   be   considered    as 
composed  of  two  factors:  5  and  100,     Multiplying  by       6^5 
•5,  the  product  is  8125;  and  the  product  of  this  number  500 

by  100  is  812500,  which  is  the  same  as  annexing   two     312  5  00 
<'iphcrs  to  the  first  product. 


MULTIPLICATION  OF  SIMPLE  NUMBERS. 


49 


2.  Find  the  product  of  2300  X  170. 

2300 
Analysis. — The   number  2300  may  be   regarded  as       17  0 


composed  of  the  two  factors  23  and  100;    and  170,  of    161 

the  two  factors  17  and  10.  2  3 


391000 


The  product  of  2300  by  170  will  be  found  by  multi- 
plying 23  by  17,  and  this  product  by  100,  and  the 
resulting  product  by  10  (33);  that  is,  by  finding  the 
product  of  23  by  17,  and  then  annexing  3  ciphers  to 
the  product,  as  there  tire  3  ciphers  at  the  right  of 
both   factors. 

Rule. — Multiply  without  regarding  the  ciphers  on  the  right 
of  the  factors ;  then  annex  to  the  product  as  many  ciphers 
as  are  at  the  right  of  both  factors. 


3. 

Multiply 

2350  by  60. 

141000. 

4. 

Multiply 

80300  by  450. 

36135000. 

5. 

Multiply 

10240  by  3200. 

32768000, 

6. 

Multiply 

9600  by  2400. 

23040000. 

7. 

Multiply 

18001  by  26000. 

468026000, 

8. 

Multiply 

8602  by  1030. 

8860060. 

9. 

Multiply 

3007  by  9100. 

27363700, 

10. 

Multiply 

80600  by  7002. 

564361200. 

11. 

Multiply 

70302  by  80300. 

5645250600. 

12. 

Multiply 

904000  by  10200. 

9220800000, 

m 


..gmm^     ^i^ 


35.  1.  If  you  divide  6  apples  equally  between  2  boys, 
how  many  will  each  boy  have? 

Analysis. — It  will  require  2  apples  to  give  each  boy  1.  Hence, 
each  boy  will  have  as  many  apples  as  2  is  contained  times  in  (i, 
which  are  3. 

How  many  times  2  in  6?  Ans.  3. 

Why?  Because  3  times  2  are  6. 

2.  If  you  divide  8  peaches  equally  between  2  boys, 
how  many  will  each  have?  Ans.  4  peaches.     Why? 

3.  How  many  times  2  in  10?  Ans.  5.     Why? 
The    process    by    which    the    preceding    examples    arc 

solved  is  called  Division. 


DEFINITIONS. 

36.  1.  Division  is  the  process  of  finding  how  many 
times  one  number  is  contained  in  another. 

2.  The  divisor  is  the  number  by  which  to  divide ;  the 
dividend  is  the  number  to  be  divided ;  the  quotient  is 
the  number  denoting  how  many  times  the  divisor  is 
contained   in  the  dividend. 


DIVISION  OF  SIMPLE  NUMBERS.  51 

Thus,  3  is  contained  in  12,  4  times;  here,  3  is  the  divisor,  12  the 
dividend,  and  4  the  quotie7it. 

3.  Since  3  is  contained  in  12  four  times,  4  times  3 
are  12;  that  is,  the  divisor  and  quotient  multiplied  pro- 
duce the  dividend. 

4.  Since  3  and  4  are  factors  of  the  product  12,  the 
divisor  and  quotient  correspond  to  the  factors  in  multi- 
plication ;  the  dividend,  to  the  product.  Therefore,  Bi- 
insion  is  the  process  of  finding  one  of  the  factors  of  a 
product,  when   the  other  factor  is  known. 

37.  A  boy  has  8  cents :  how  many  lemons  can  he  buy, 
at  2  cents  each? 

Analysis. — He  can  buy  4,  because  4  lem-  8  cents, 

ons,  at  2  cents  each,  will  cost  8  cents.  1st  lemon,  2  cents. 

The  boy  would  give  2  cents  for  1  lemon,  Left,             G  cents, 

and  then  have  6  cents  left.  2d    lemon,  2  cents. 

After  giving  2  cents  for  the  2d  lemon,  he  Left,            4  cents, 

would  have  4  cents  left.  3d    lemon,  2  cents. 

Then,  giving  2  cents  for  the  3d,  he  would  Left,             2  cents, 

have  2  cents  left.  4th  lemon,  2  cents. 

Lastly,  after  giving  2  cents  for  the  4th,  he  Left,  0  cents, 
would  have  nothing  left. 

The  natural  method  of  performing  this  operation  is  by 
subtraction;  but,  when  it  is  known  how  many  times  2 
can  be  subtracted  from  8,  instead  of  subtracting  2  four 
times,  say  2  in  8  four  times,  and  4  times  2  are  8. 

Therefore,  Division  may  be  termed  a  short  method  of 
making  many  subtractions  of  the  same  number. 

The  divisor  is  the  number  subtracted ;  the  dividend, 
the  number  from  which  the  subtraction  has  been  made ; 
the  quotient  shows  how  many  subtractions  have  been 
made. 


52 


KAY'S  NEW  PKACTICAL  ARITHMETIC. 


38.     1.  Division  is  indicated  in  three  ways: 

1st.    3)12,    wliieb  means  that  12  is  to  be  divided  by  3. 

12 
2d.        ^       which  means  that  12  is  to  be  divided  by  3. 

D 

3d.    12-^3,  which  means  that  12  is  to  be  divided  by  3. 

2.  In  using  the  first  sign  wiien  the  divisor  does  not 
exceed  12,  draw  a  line  under  the  dividend,  and  write 
the  quotient  beneath;  if  the  divisor  exceeds  12,  draw  a 
curved  line  on  the  right  of  the  dividend,  and  place  the 
quotient  on  the  right  of  this. 

3.  The  sign  (-^)  is  read  divided  by. 


Examples. 


2)8 
4 


15)45(3 
45 


¥-« 


21-^3=7. 


Division  Table. 


1-5-1=  1 

2-^2=  1 

8--3^  1 

4-4=  1 

2--l=  2 

4h^2=  2 

6--8=  2 

8-j-4=  2 

3--l=  8 

6 --2=  8 

9--3=  3 

12^4=^  3 

4--l==  4 

8-^2=  4 

12-:~8=  4 

16-^4=  4 

5-r-lr^  5 

10-1-2=  5 

15  H- 3=  b 

20^4=  6 

0^1^  G 

12^2=  6 

l8--8=:  6 

24-^4=  6  1 

7-^-1=  7 

14  -r-  2=  7 

21  -  3  =  7 

28-4-4=  7 

8--l=r  8 

16-2=  8 

24 -f- 3=  8 

32  4-4=  8 

9  H-  1  —  9 

18-^2=  9 

27-4-8=  9 

86^4=  9 

10--1==10 

20 --2  =  10 

80  -^  3  =  10 

40-4-4  =  10 

11--1==11 

22  H-  2  =  1 1 

83 ---3  =  11 

44-4-4  =  11 

12 --1=12 

24 -=-2  =  12 

36-3  =  12 

48 -r- 4  =  12 

DIVISION  OF   SIMPLE   NUMBERS. 


53 


1 

5--5=r   1 

6  —  6=  1 

7  —  7=  1 

8- 

-8=  1 

10 --5=  2 

12  —  6=  2 

14-^7=,  2 

16- 

-8=  2 

15^5=  3 

18  —  6=  3 

21  —  7=  3 

24- 

-8=  3 

20 -f- 5==  4 

24  —  6=  4 

28  —  7=  4 

32- 

-8=  4 

25^-5=  5 

30-^6=  5 

35  —  7=  5 

40- 

-8=  5 

30 --5=  6 

3G^6=  6 

42^7=  6 

48- 

-8=  6 

S5-^6=    7 

42  —  6=  7 

49  —  7=  7 

56- 

-8=7 

40-v-5=:r  8 

48  —  6=  8 

56  —  7=  8 

64- 

-8=  8 

45--5=?  9 

54  —  6=  9 

63  —  7=  9 

72- 

-8=  9 

50 --5  =  10 

60  —  6  =  10 

70  --  7  =  10 

80- 

-8  =  10 

55-5  =  11 

66  —  6  =  11 

77  —  7  =  11 

88- 

-8  =  11 

60 --5  =  12 

72  —  6  =  12 

84  —  7  =  12 

96- 

-8  =  12 

9--9=  1 

10  —  10=  1 

11—11=  1 

12- 

-12=  1 

18-9=  2 

20- 

-10=  2 

22  —  11=  2 

24- 

-12=  2 

27  —  9=  3 

30- 

-10=  3 

33_^11^  3 

36- 

-12=  3 

36  —  9=  4 

40- 

-10=  4 

44_^11^  4 

48- 

-12=  4 

45  _^  9=:  5 

50- 

-10=  5 

55  —  11=  5 

60- 

-12=  5 

54_j_.9=  6 

60- 

-10=  6 

66  —  11=  6 

72- 

-12=  6 

63  —  9=  7 

70- 

-10=  7 

77  —  11=  7 

84- 

-12=  7 

72  —  9=  8 

80- 

-10=  8 

88  —  11=  8 

96- 

-12=  8 

81  —  9  =  9 

90- 

-10=  9 

99  —  11=  9 

108- 

-12=  9 

90-9  =  10 

100- 

-10  =  10 

110  —  11=10 

120- 

-12  =  10 

99  —  9  =  11 

110- 

-10  =  11 

121  —  11=11 

132- 

-12=11 

108  —  9  =  12 

120- 

-10  =  12 

132  —  11=12 

144- 

-12  =  12 

39.  If  7  cents  be  divided  as  equally  as  possible  among 
3  boys,  each  boy  would  receive  2  cents,  and  there  would 
be  1  cent  left,  or  remaining  undivided. 

The  number  left  after  dividing,  is  called  the  re- 
raainder. 

Rem. — 1.  Since  the  remainder  is  a  part  of  the  dividend,  it  must  be 
of  the  same  denomination.  If  the  dividend  he  dollars,  the  remainder 
will  he  dollars;  if  pounds,  the  remainder  will  he  pounds. 


54  RAY'IS  NEW  PRACTICAL  ARITHMETIC. 

Rem.  2.— The  remainder  is  always  lens  than  the  divisor;  for,  if  it 
were  equal  to  it,  or  greatei*  the  divisor  would  be  contained  at  least 
once  more  in  the  dividend. 

Rem.  3. — If  the  dividend  and  divisor  are  simple  numbers,  the 
operation  is  called  Division  of  Simple  Numbers. 


Short  Division. 

40.  When  the  division  is  performed  mentallj^,  and 
merely  the  result  written,  it  is  termed  Short  Division. 
Short  Division  is  used  when  the  divisor  does  not  exceed 
12. 

1.  How  many  times  is  2  contained  in  468? 

Here,  the  dividend  is  composed  of  three  numbers;  4  hundreds,  6 
tens,  and  8  units;  that  is.  of  400,  60,  and  8. 

Divisor.     Dividend.  Quotient. 

Now,       2     in     400     is  contained     200     times. 

2     in       60     ''  ''  30     times. 

2     in         8     ^'  "  4     times. 

Hence,    2     in     468     is  contained     234     times. 

The  same  result  can  be  obtained  without  actually 
separating  the  dividend  into  parts : 

Thus,  2  in  4  (hundreds),  2  times,  which  write  Dividend, 

in  hundreds'  place;  then,  2  in  6  (tens),  3  times.  Divisor,  2)468 

which  write  in  tens'  place;  then,  2  in  8  (units),  4  Quotient,     2  34 
times,  which  write  in  units'  place. 

2.  How  many  times  3  in  693?  231. 

3.  How  many  times  4  in  848?  212. 

4.  How  many  times  2  in  4682?  2341. 

5.  How  many  times  4  in  8408?  2102. 


DIVISION  OF  SIMPLE  NUMBERS.  55 

6.  How  many  times  3  in  3693G?  12312. 

7.  How  many  times  2  in  88468?  44234. 

41.     1.  How  many  times  is  3  contained  in  129? 

Solution. — Here,  3  is  not  contained  in  1 ;   but  3  is       3)129 
contained  in  12   (tens),  4  times,   which  write  in  tens'  4  3 

place;  3  in  9  (units),  3  times,  which  write  in  units'  place. 

2.  How  many  times  is  3  contained  in  735? 

Solution. — Here,  3  is  contained  in  seven  (hundreds), 
2   times,  and   1    hundred  over;    the  1  hundred,  united        3)735 
with  the  3  tens,  makes  13  tens,  in  which  3  is  contained  245 

4   times  and   1  ten  left;    this  1  ten,  united  with  the  5 
units,  makes  15  units,  in  which  3  is  contained  5  times. 

3.  How  many  times  is  3  contained  in  618? 

SoLUTiON.-r-Here,  3  is  contained  in  6  (hundreds),  2 
times;  as  the  1  in  ten's  place  will  not  contain  3,  a  cipher        3)618 
IS  placed  in  ten's  place;  the  1  ten  is  then  added  to  the  8  206 

units,  making    18  units,  and   the  quotient   figure  6  is 
placed  in  units'  place. 

4.  How  many  times  is  3  contained  in  609? 

Here,  the   solution   is   the  same  as  in  the  above  ex-        3)609 
ample;  there  being  no  tens,  their  order  is  indicated  by  0.  20  3 

5.  How  many  times  is  3  contained  in  743? 

After  dividing,  there  is  2  left,  the  division  of  which 
is  merely  indicated  by  placing  the  divisor  under  the         3)743 
remainder;    thus,   f.     The   quotient   is  written   thus,  24  7§ 

247f;   read,  247,   and  two  divided  by  three;  or,  247, 
with  a  re7nainder,  two. 


56  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

6.  How  many  times  3  in  462?  154. 

7.  How  many  times  5  in  1170?  234. 

8.  How  many  times  4  in  948?  237. 

Rule. — 1.  Wiite  the  divisor  at  the  left  of  the  dividend, 
tvith  a  curved  line  between  them^  and  draw  a  line  beneath 
the  dividend.  Begin  at  the  left  hand,  divide  successively 
each  figure  of  the  dividend  by  the  divisor,  and  write  the  re- 
sult in  the  same  order  in  the  quotient. 

2.  If  there  is  a  remainder  after  dividing  any  figure, 
'prefix  it  to  the  figure  in  the  next  lower  order,  and  divide 
as  before. 

3.  If  the  number  in  any  order  does  not  contain  the  divisor, 
place  a  cipher  in  the  same  order  in  the  quotient,  prefix  the 
number  to  the  figure  in  the  next  lower  order,  and  divide  as 
before. 

4.  If  there  is  a  remainder  after  dividing  the  last  figure, 
place  the  divisor  under  it,  and  annex  it  to  the  quotient. 

Proof. — Multiply  the  quotient  by  the  divisor,  and  add 
the  remainder,  if  any,  to  the  product:  if  the  work  is 
correct,  the  sum  will  be  equal  to  the  dividend. 

Rem. — This  method  of  proof  depends  on  the  principle  (36,  4] 
that  a  dividend  is  a  product,  of  which  the  divisor  and  quotient  are 
factors. 

9.  Divide  653  cents  by  3. 

SOLUTION.  TROOF. 

217 

Dividend.  3 

Divisor,     3)653  (151=  cents  divided. 

Quotient,        2 1  7  §  2  =  remainder. 

¥53  =  dividend. 


DIVISION  OF  SIMPLE  NUMBEKS. 


57 


(10) 
6)454212 

(11) 

7)874293 

(12) 
8)3756031 

Ans. 

75702 
6 

124899 

7 

469503J 

8 

Proof, 

454212 

874293 

3756031 

PARTS  OF    NUMBERS. 

Note. — When  any  number  is  divided  into  two  equal  parts,  one  of 
the  parts  is  called  one-half  of  that  number. 

If  divided  into  three  equal  parts,  one  of  the  parts  is  called  07ie- 
third;  if  into  four  equal  parts,  one-fourth;  if  into  live  equal  parts, 
one-fifth;  and  so  on. 

Hence,  to  find  one-half  of  a  number,  divide  by  2;  to  find  one-third^ 
divide  by  3;  one-fourth,  divide  by  4;  one-ffth,  by  5,  etc. 


4326. 

13541  If. 

1687601 . 

196855. 

4311 7^. 

1234753f 

754065. 

1003634. 

1830023-V 

54841. 

3472834. 

24.  If  oranges    cost    3    cents    each,  how  many  can  be 
bought  for  894  cents?  298.. 

25.  If  4  bushels  of  apples  cost  140  cents,  how  much  is 
that  a  bushel?  35  ct. 

26.  If  flour  cost  84  a  barrel,  how  many  barrels  can  be 
bought  for  $812?  203. 


13. 

Divide 

8652  by 

2. 

14.  -IHvide 

406235  by 

3. 

15. 

Divide 

675043  by 

4. 

16. 

Divide 

984275  by 

5. 

17. 

Divide 

258703  by 

6. 

18. 

Divide 

8643275  by 

7. 

19. 

Divide 

6032520  by 

8. 

20. 

Divide 

9032706  by 

9. 

21. 

Divide 

1830024  by 

10. 

22. 

Divide 

603251  by 

11. 

23. 

Divide  41674008  by 

12. 

58  liAY'S  NEW  FKAGTICAL  ARITHMETIC. 

27.  A  carpenter  receives  $423  for  9  months'  work : 
liow  much  is  that  a  month?  $47. 

28.  There  are  12  months  in  1  year:  how  many  years 
are  there  in  540  months?  45. 

29.  There  are  4  quarts  in  1  gallon :  how  many  gallons 
are  there  in  321276  quarts?  80319. 

30.  At  $S  a  barrel,  how  many  barrels  of  flour  can  be 
bought  for  $1736?  217. 

31.  There  are  7  days  in  one  wxek :  how  many  weeks 
are  there  in  734566  days?  "      .  104938. 

32.  A  number  has  been  multij^lied  by  11,  and  the  i)ro- 
duct  is  495  :  what  is  the  number  ?  45. 

33.  The  product  of  two  numbers  is  3582 :  one  of  the 
numbers  is  9  :  what  is  the  other  ?  398. 

34.  Find  one-half  of  56.  28. 

35.  Find  one-half  of  3725.  '  1862^. 

36.  Find  one-third         of  147.  49. 

37.  Find  one-fourth       of  500.  125. 

38.  Find  one-fifth  of  1945.  389. 

39.  Find  one-sixth         of  4476.  746. 

40.  Find  one-seventh     of  2513.  .^  359. 

41.  Find  one-eighth       of  5992.  *  749. 

42.  Find  one-ninth        of  8793.  977. 

43.  Find  one-tenth         of  1090.  r  109. 

44.  Find  one-eleventh  of  4125.  375. 

45.  Find  one-twelfth      of  5556.  463. 

46.  I  divided  144  apples  equally  among  4  boys;  the 
eldest  boy  gave  one-third  of  his  share  to  his  sister :  what 
number  did  the  sister  receive?  12. 

47.  James  found  195  cents,  and  gave  to  Daniel  one- 
fifth  of  them:  Daniel  gave  one-third  of  his  share  to  his 
sister:  how  many  cents  did  she  receive?  13. 

48.  One-eleventh  of  275  is  how  much  greater  than 
one-eighth  of  192?  1. 


DIVISION  OF  SIMPLE  NUMBERS.  59 


Long  Division. 

42,  When  the  entire  work  of  the  division  is  written 
down,  it  is  termed  Long  Division.  Long  Division  is 
commonly  used  when  the  divisor  exceeds  12. 

1.  Divide  3465  dollars  equally  among  15  men. 

Solution. — Fifteen   is   not  contained    in    3 
(thousands) ;   therefore,  there  will  be  no  thou-       15)3465(231 
sands  in  the  quotient.     Take  34  (himdreds)  as  3  0  hund. 

Si  partial   dividend;    15  is  contained  in  34,  2  4  6  tens, 

times;  that  ig,  15  men  have  200  dollars  each,  4  5 

which  requires  in  all  15  X  2  =  30  hundreds  of  15  units, 

dollars.  1 5 

Subtract  30  hundreds  from  34  hundreds,  and 
4  hundreds  remain;  to  which  bring  down  the  6  tens,  and  you  have 
46  (tens)  for  a  second  partial  dividend. 

46  contains  15,  3  times;  that  is,  each  man  has  30  dollars  more,  and 
all  require  15  X  ^  =^  45  tens  of  dollars. 

Subtract  45,  and  bring  down  the  5  units,  which  gives  15  (units)  for 
a  third  partial  dividend;  in  this  the  divisor  is  contained  once,  giving 
to  each  man  1  dollar  more. 

Hence,  each  man  receives  2  hundred  dollars,  3  ten 
dollars,  and  1  dollar;  that  is,  231  dollars. 

By  this  process,  the  dividend  is  sep- 
arated into  parts,  each  part  contain-      Divisor.     Parts.     Quotients 
ing  the  divisor  a  certain  number  of  15  3000         200 

times.  4  5  0  3  0 

The   first    part,    30   hundreds,    con-  15  1 

tains  the  divisor  2   times;  the  second  3  4  65  231 

part,  45  tens,  contains  it  3  times;  the 
third  part,  15  units,  contains  it  1  time. 

The  several  parts  together  equal  the  given  dividend,  and  the 
several  partial  quotients  make  up  the  entire  quotient. 


60  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  In  147095  days,  how  many  years,  each  of  365  days? 

Solution.— Taking    147    (thou-    865  )  1  4  7005  (  403  years, 
sands)  for  the  first  partial  dividend,  14  60 

we  find  it  will  not  contain  the  di-  109  5 

visor;  hence  use  four  figures.  109  5 

Again,  after  multiplying  and  sub- 
tracting, as  in  the  preceding  example,  and  bringing  down  the  9  tens, 
the  partial  dividend,  109  (tens),  will  not  contain  the  divisor;  hence, 
write  a  cipher  (no  tens)  in  the  quotient,  and  bring  down  the  5  units; 
the  last  partial  dividend  is  1095  (units),  which  contains  the  divisor 
three  times. 

3.  Divide  4056  by  13.  312. 

Rule. — 1.  Flace  the  divisor  on  the  left  of  the  dividend^ 
draw  a  curved  line  between  them,  and  another  on  the  right  of 
the  dividend. 

2.  Find,  how  many  times  the  divisor  is  contained  in  the 
fewest  left  hand  figures  of  the  dividend  that  will  contain 
the  divisor^  and  place  this  number  in  the  quotient  at  the 
right. 

3.  Multiply  the  divisor  by  this  quotient  figure;  place  the 
^product  under  that  part  of  the  dividend  from  which  it  was 
obtained. 

4.  Subtract  this  product  from  the  figures  above  it ;  to  the 
remainder  bring  down  the  next  figure  of  the  dividend,  and 
divide  as  before,  until  all  the  figures  of  the  dividend  are 
brought  down. 

'  5.  If  at  any  time,  after  bringing  down  a  figure,  the  num- 
ber thus  formed  is  too  smcdl  to  contain  the  divisor,  place  a 
cipher  in  the  quotient,  and  bring  down  another  figure,  after 
which  divide  as  before. 

Proof. — Same  as  in  Short  Division. 


DIVISION  OF  SIMPLE  NUMBERS. 


61 


Rem. — 1.  The  product  must  never  be  gr eater  than  the  partial 
dividend  from  which  it  is  to  be  subtracted;  when  so,  the  quotient 
figure  is  too  large,  and  must  be  diminished. 

Rem. — 2.  After  subtracting,  the  remainder  must  always  be  less 
than  the  divisor;  when  the  remainder  is  not  less  than  the  divisor,  the 
last  quotient  figure  is  too  S7nall,  and  must  be  increased. 

Rem. — 3.  The  order  of  each  quotient  figure  is  the  same  as  the 
lowest  order  in  the  partial  dividend  from  which  it  was  obtained. 

4.  Divide  78994  by  319. 


SOLUTION 

PROOF. 

3  1  9  )  7  8  9  9  4  ( 

247Hi 

247 

Quotient 

638 

319 

Divisor. 

1519 

2223 

12  76 

247 

2434 

741 

2233 

78793 

201 

Rem. 

Add 

201 

78994 

Remainder. 
=  the  Dividend. 

5.  Divide 

11577 

by 

14. 

-.«26|f. 

6.  Divide 

48690 

by 

15. 

3246. 

7.  Divide 

1110960 

by 

23. 

48302^1. 

8.  Divide 

122878 

by 

67. 

1834. 

9.  Divide 

12412 

by 

53. 

234if. 

10.  Divide 

146304 

by 

72. 

2032. 

11.  Divide 

47100 

by 

54. 

872i|. 

12.  Divide 

71104 

by 

88. 

808. 

13.  Divide 

43956 

by 

66. 

666. 

14.  Divide 

121900 

by 

99. 

1231ff 

15.  Divide 

25312 

by 

112. 

226. 

16.  Divide 

381600 

by 

123. 

3102A\. 

17.  Divide 

105672 

by 

204. 

518. 

18.  Divide 

600000 

by 

1234. 

486^3V 

19.  Divide 

1234567 

by 

4321. 

285itff 

20.  Divide 

50964242 

by 

7819. 

6518. 

62  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

21.  Divide     48905952  by  9876.  4952. 

22.  Divide       4049160  by  12345.  328. 

23.  Divide  552160000  by  973.  567482^VV 

24.  At  $15  an  acre,  how  many  acres  of  land  can  be 
bought  for  $3465?  .  231  acres. 

25.  If  a  man  travel  26  miles  a  day,  in  how  many  days 
will  he  travel  364  miles?  14  days. 

26.  If  $1083  be  divided  equally  among  19  men,  how 
many  dollars  will  each  have?  $57. 

27.  A  man  raised  9523  bushels  of  corn  on  107  acres: 
how  much  was  that  on  one  acre?  89  bu. 

28.  In  1  hogshead  there  are  63  gallons :  how  many 
hogsheads  in  14868  gallons?  236. 

29.  The  President  receives  $50000  a  year  (365  days): 
how  much  is  that  a  day?  $136  and  $360  over. 

30.  The  yearly  income  from  a  railroad  is  $379600 : 
how  much  is  that  a  day?     (365  da.  =:=  1  yr.)  $1040. 

31.  The  product  of  two  numbers  is  6571435 ;  one  of 
the  factors  is  1235:  what  is  the  other?  5321. 

32.  Divide  one  million  two  hundred  and  forty-seven 
thousand  four  hundred  by  405.  3080. 

33.  Divide  10  million  four  hundred  and  one  thousand 
by  one  thousand  and  six.  10338yyg?_ 

34.  A  colony  of  684  men  bought  a  tract  of  land,  con- 
taining 109440  acres :  if  equally  divided,  to  how  many 
acres  was  each  man  entitled?  160  acres. 

35.  A  farmer  raised  8288  bushels  of  corn,  averaging 
56  bushels  to  the  acre:  how  many  acres  did  ho  plant? 

148  acres. 

36.  The  capital  of  a  joint-stock  company  is  $262275, 
and  is  divided  into  269  shares :  what  is  the  value  of 
each  share?  $975. 

37.  The  earth,   at  the  equator,  is   about  24899  miles*  in 


DIVISION  OF  SIMPLE  NUMBERS.  63 

circumference,  and    turns    on    its  axis  once  in  24  hours : 
how  many  miles  an  hour  does  it  turn?  1037^^. 

38.  A  railroad  238  miles  long,  cost  $3731840:  what 
was  the  cost  per  mile?  $15680. 

39.  A  fort  is  27048  feet  distant  from  a  city;  the  flash 
of  a  cannon  w^as  seen  24  seconds  before  the  sound  was 
heard :   how  many  feet  a  second  did  the  sound  travel  ? 

1127  feet. 

40.  Light  travels  at  the  rate  of  11520000  miles  a  min- 
ute :  how  many  minutes  does  it  require  for  the  light  of 
the  sun  to  reach  the  earth,  the  sun  being  92160000  miles 
distant?  8  minutes. 

Examples  for  TIkvtew. 

41.  Subtract  86247  from  94231  and  divide  the  re- 
mainder by  16.  499. 

42.  Divide  the  sum  of  46712  and  6848  by  104.        515. 

43.  Divide  the  product  of  497  X  583  by  71.  4081. 

44.  To  the  difference  between  2832  and  987  add  678, 
and  divide  the  sum  by  87.  29. 

45.  Multipl}^  the  diflPerence  between  4896  and  2384  by 
49,  and  divide  the  product  by  112.  1099. 

46.  Multiply  the  sum  of  228  +  786  by  95,  and  divide 
the  product  by  114.  845. 

47.  Multiply  the  sum  of  478  and  296  by  their  differ- 
ence, and  divide  the  product  by  387.  364. 

48.  A  horse-dealer  received  S7560  for  horses ;  he  sold 
a  part  of  them  for  S3885 ;  if  he  sold  the  rest  for  $175 
apiece,  how  many  horses  did  he  sell  the  second  time? 

21  horses. 

49.  A  farmer  expended  at  one  time  $7350  for  land,  and 
at  another,  $4655,  paying  $49  an  acre  each  time :  how 
many  acres  did  he  buy  in  both  purchases?         245  acr^. 


64  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

50.  A  refiner  bought  58  hogsheads  of  sugar,  at  $77 
a  hogshead,  and  afterward  sold  them  for  $5742:  how 
much  did  he  gain  on  each  hogshead?  $22. 

51.  A  man  bought  240  acres  of  land,  at  $26  an  acre, 
giving  in  payment  a  house  valued  at  $2820,  and  horses 
valued  at  $180  apiece:  how  many  horses  did  he  give? 

19  horses. 

52.  A  speculator  bought  25  acres  of  land  for  $10625, 
and  after  dividing  it  into  125  village  lots,  sold  each  lot 
for  $250:  how^  much  did  he  gain  on  the  whole?  On  each 
acre?     On  each  lot?  $20625.     $825.     $165. 


CONTRACTIONS  IN  DIVISION. 
CASE    I. 

43.     When  the  divisor  can  be  separated  into  factors. 

1.  A  man  paid  $255  for  15  acres  of  land:  how  much 
was  that  per  acre? 

Solution. — 15    acres    are    3  Dollars, 

times  5  acres;  dividing  $255  by        3)255  =  the  value  of  15  acres. 
3    gives    $85,   the   value    of    5  6)85  —  the  value  of    6  acres, 

acres;   dividing  $85  by  6  gives  1  7  =  the  value  of    1  acre. 

$17,  the  value  of  1  acre. 

The  solution  shows  that  instead  of  dividing  by  the  number  15, 
whose  factors  are  3  and  5,  we  may  first  divide  by  one  factor,  then 
divide  the  quotient  thus  obtained  by  the  other  factor. 

2,  Find  the  quotient  of  37,  divided  by  14. 

Solution. — Dividing  by  2,  the  quotient 
is  18  twos  and  1  unit  remaining.     Divid-     2)37 
ing  by  7,  the  quotient  is  2,  with   a  re-     7)18  and  1  over, 
mainder  of  4  tioos;  the  whole  remainder        ^^  and  4  twos  left. 
then  is  4  ttcos  plus  1,  or  9. 


DIVISION  OF  simplp:  numbers.  65 

Rule. — 1.  Divide  the  dividend  by  one  of  the  factors  of 
the  divisor;  then  divide  the  quotient  thus  obtained  by  the 
other  factor. 

2.  Multiply  the  last  remainder  by  the  first  divisor ;  to  the 
product  add  the  first  remainder ;  the  amount  ivill  be  the  true 
remainder. 

Rem. — When  the  divisor  can  be  resolved  into  more  than  two 
factors,  you  may  divide  by  them  successively.  The  true  remainder 
will  be  found  by  multiplying  each  remainder  by  all  the  preceding 
divisors,  except  that  which  produced  it.  To  their  sum  add  the  re- 
mainder from  first  divisor. 

3.  Divide     2583  by  63. 

4.  Divide     6976  by  32. 

5.  Divide     2744  by  28. 

6.  Divide     6145  by  42. 

7.  Divide  19008  by  132. 

8.  Divide     7840  by  64. 

9.  Divide  14771  by  72. 

10.  Divide  10206  by  81. 

11.  Divide  81344  by  121. 

12.  Divide  98272  by  108. 

CASE    II. 

44.  To  divide  bj^  1  with  ciphers  annexed;  as  10,  100, 
iOOO,  etc. 

To  multiply  6  by  10,  annex  one  cipher,  thus,  60.  On 
the  principle  that  division  is  the  reverse  of  multiplica- 
tion, to  divide  60  by  10,  cut  off  a  cipher. 

Had  the  dividend  been  65,  the  5  might  have  been  separated  in  like 
manner  as  the  cipher;  6  being  the  quotient,  5  the  remainder.     The 
sam-;  will  apply  when  the  divisor  is  100,  1000,  etc. 
Prac-  5 


(t)3  = 

=  7X9) 

41. 

(32  = 

=  4X8) 

218. 

(28  = 

=  7X4) 

98. 

(42  = 

=  6X7) 

146if. 

144. 
122ff. 
205||. 

126. 
672/A. 

66 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


Rule.—  Cut  off  as  viany  figures  from  the  right  of  the 
dividend  as  there  are  ciphers  in  the  divisor ;  the  figures  cut 
off  will  be  the  remainder,  the  other  figures,  the  quotient. 

1.  Divide  34872  by  100. 


1|00 

Ol'KR 

)348| 

ATIO.V. 

72 
Quo. 

72 

Rem 

348 

2^ 
3. 
4. 
5. 
6. 

Divide 
Divide 
Divide 
Divide 
Divide 

2682  by  10. 

4700  by  100. 
37201  by  100. 
46250  by  100. 
18003  by  1000. 

268ft. 

47. 

372t^. 

462AV 

CASE    ITT. 

45.     To  divide  when  there  are  ciphers  on  the  right  of 
the  divisor,  or  on  the  right  of  the  divisor  and  dividend. 


1.  Divide  4072  by  800. 

Solution.  —  Regard  800  as  com- 
posed of  the  factors  100  and  8,  and 
divide  as  in  the  margin. 

In  dividing  by  800,  separate  the 
two  right  hand  figures  for  the  re- 
mainder,  then    divide   by  8. 

2.  Divide  77939  by  2400. 


OPERATION. 


I|00)40i72 


8^ 


40 
5  Quo... 72  Rem. 


8100)40172 

5  Quo... 72  Rem. 


Solution.  —  Since  2400  equals 
24  X  100,  cut  off  the  two  right  hand 
figures,  the  same  as  dividing  by  100; 
then  divide  by  24. 

Dividing  by  100,  the  remainder  is 
39;  dividing  by  24,  the  remainder  is  11. 

To  find  the  true  remainder,  multiply  11  by  100,  and  add  39  to 
the  product  (Art.  43,  Rule);  this  is  the  same  as  annexing  the 
figures   cut   oflf,  to   the   last   remainder. 


operation. 
24100)7  79139(32  1^^1 

72 

48 


DIVISION  OF  SIMPLE  NUMBEKS. 


67 


3.  Divide  62700  by  2500. 


Solution. — The 
example    above. 


same    as    for   the 


OPERATION. 

25iOO)627|00(25//o°^ 
50 

T2T 

125 

2 


Rule. — 1.   Cut  off  the  ciphers  at  the  light  of  the  divisor^ 
and  as  many  figures  from  the  right  of  the  dividend. 

2.  Divide  the  remaining  figures  in  the   dividend   by    the 
remaining  figures  in  the  divisor. 

3.  Annex  the  figures  cut  off  to  the  remainder^  which  gives 
the  triie  remainder. 


4. 

Divide 

73005  by  4000. 

5. 

Divide 

36001  by  9000. 

6. 

Divide 

1078000  by  11000. 

7. 

Divide 

40167  by  180. 

8. 

Divide 

907237  by  2100. 

9. 

Divide 

364006  by  6400. 

10. 

Divide  ' 

76546037  by  250000. 

11. 

Divide  43563754  by  63400. 

1844>4>A 

98. 

223jV^. 

432^H^. 

306/A^V\\- 
687^Wo. 


GENERAL  PRINCIPLES  OF  DIVISION. 


46.  The  value  of  the  quotient  depends  on  the  rela- 
tive values  of  divisor  and  dividend.  These  may  be 
changed  by  Multiplication  and  by  Division,  thus : 

1st.  The  dividend  may  be  multiplied,  or  the  divisor 
divided. 

2d.  The  dividend  may  be  divided,  or  the  divisor  multi- 
plied. 

3d.  Both  dividend  and  divisor  may  be  multiplied,  or 
both  divided,  at  the  same  time. 


68  RAY'S  NEAV  PRACTICAL  ARITHMETIC. 

Illustrations. 

Let  24  be  a  dividend,  and  6  the  divisor;  the  quotient 
is  4.     24-^6  =  4. 

If  the  dividend,  24,  be  multiplied  by  2,  the  quotient  will  be  multi- 
plied by  2;  for,  24X2  =  48;  and  48-v-6=:8,  which  is  the  former 
quotient,  4,  multiplied  by  2. 

Now,  if  the  divisor,  6,  be  divided  by  2,  the  quotient  will  be  multi- 
plied by  2;  for,  6 --2  =  3;  and  24-f-3  =  8,  which  is  the  former 
quotient,  4,  multiplied  by  2. 

Principle  I. — If  the  dividend  be  multiplied,  or  the  divisor 
be  divided,  the  quotient  will  be  multiplied. 

47.  Take  the  same  example,  24  -f-  6  =  4. 

If  the  dividend,  24,  be  divided  by  2,  the  quotient  will  be  divided 
by  2;  for  24 -=-2  =  12;  and  12-h6  =  2,  which  is  the  former  quo- 
tient, 4,  divided  by  2. 

And,  if  the  divisor,  6,  be  nudtiplied  by  2,  the  quotient  will  be 
divided  by  2;  for,  6  X  2  =  12;  and  24  h-  12  =  2,  which  is  the  former 
quotient,  4,  divided  by  2. 

Prin.  II. — If  the  dividend  be  divided,  or  the  divisor  be 
mxdtiplied,,  the  quotient  will  be  divided. 

48.  Take  the  same  example,  24  -f-  6  =  4. 

If  the  dividend,  24,  and  divisor,  6,  be  multiplied  by  2,  the  quotient 
will  not  be  changed;  for,  24  X  2  =  48 ;  and  6X2  =  12;  48^-12  =  4; 
the  former  quotient,  4,  unchanged. 

And  if  the  dividend,  24,  and  divisor,  6,  be  divided,  by  2,  the  quo- 
tient will  not  be  changed;  for,  24 -f- 2  =  12;  and  6-f-2  =  3;  12^3 
-=4;  the  former  quotient,  4,  unchanged. 

Prin.  III.^ — If  both  dividend  and  divisor  be  jmdtiplied  or 
divided  by  the  same  number,  the  quotient  will  not  be  changed. 


DIVISION  OF  SIMPLE  NUMBERS.  69 

Promiscuous  Examples. 

49.  1.  In  4  bags  are  $500;  in  the  first,  $96;  in  the 
second,  $120 ;  in  the  third,  $55 :  what  sum  in  the  4th 
bag?  $229. 

2.  Four  men  paid  $1265  for  land :  the  first  paid  $243 ; 
the  second  $61  more  than  the  first;  the  third  $79  less 
than  the  second:  how  much  did  the  fourth  man  pay? 

$493. 

3.  1  have  five  apple  trees:  the  first  bears  157  apples; 
the  second,  264;  the  third,  305;  the  fourth,  97;  the  fifth, 
123 :  I  sell  428,  and  186  are  stolen :  how  many  apples 
are  left?  332. 

4.  In  an  army  of  57068  men,  9503  are  killed;  586 
join  the  enemy;  4794  are  prisoners;  1234  die  of  wounds; 
850  are  drowned:    how  man}-  return?  40101. 

5.  On  the  first  of  the  year  a  speculator  is  worth 
$12307 :  during  the  year  he  gains  $8706 ;  in  January  he 
sj^ends  $237  ;  in  February,  $301 ;  in  each  of  the  remain- 
ing ten  months  he  spends  $538:  how  much  had  he  at 
the  end  of  the  year?  $15095. 

6.  The  Bible  has  31173  verses:  in  how  many  days  can 
I  read  it,  by  reading  86  verses  a  day?  362|^. 

7.  I  bought  28  horses  for  $1400:  3  died;  for  how 
much  each  must  I  sell  the  rest  to  incur  no  loss?        $56. 

8.  How  many  times  can  I  fill  a  15  gallon  cask,  from 
5  hogsheads  of  63  gallons  each?  21  times. 

9.  A  certain  dividend  is  73900;  the  quotient  214;  the 
remainder,  70:  what  is  the  divisor?  345. 

10.  Multiply  the  sum  of  148  and  56  by  their  difier- 
ence ;   divide  the  product  by  23.  816. 

11.  How  much  woolen  cloth,  at  $6  a  yard,  will  it  take 
to  pay  for  8  horses  at  $60  each,  and  14  cows  at  $45 
each?  185  yd. 


70  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

12.  Two  men  paid  $6000  for  a  farm:  one  man  took 
70  acres  at  S30  an  acre,  tlie  other  the  remainder  at  $25 
an  acre  :    how  many  acres  in  all  ?  226. 

13.  My  income  is  $1800  a  year.  If  I  spend  $360  a 
year  for  provisions,  $300  for  rent,  $150  for  clothing, 
$100  for  books,  and  $90  for  incidentals,  in  how  many 
years  can  I  save  $10400?  13. 

14.  A  man  bought  40  acres  of  ground  at  $15  an 
acre,  and  80  acres  at  $25  an  acre.  He  sold  90  acres 
for  $4500,  and  the  remainder  at  $60  an  acre :  for  how 
much  did  he  sell  the  whole  land?  How  much  did  he 
gain?  $6300.     $3700. 

15.  A  merchant  bought  275  3^ards  of  cloth  at  $4  a 
yard;  he  sold  250  yards  at  $5  a  yard,  and  the  re- 
mainder at  $6  a  yard :   how  much   did  he  gain?     $300. 

16.  A  broker  buys  125  shares  of  stock  for  $85  a 
share,  and  75  shares  at  $115  a  share.  He  invests  it  all 
in  other  stock  at  $175  a  share :  how  many  shares  does 
he  get  by  the  last  purchase?  110. 

17.  A  farmer  sends  to  a  dealer  20  horses  and  15 
mules  to  be  sold.  The  dealer  sells  the  horses  for  $125 
each,  and  the  mules  for  $150  each,  charging  $95  for 
selling.  The  farmer  then  buys  50  head  of  cattle  at  $45 
each,  with  part  of  the  money,  and  deposits  the  remainder 
in  bank:    how  much  does  he   deposit  in   bank?     $2405. 


To  Teachers.  —  While  placing  Fractions  immediately  after 
Simple  Whole  Numbers  is  philosophical,  and  appropriate  in  a 
higher  arithmetic  for  advanced  pupils,  the  experience  of  the  author 
convinces  him  that,  in  a  book  for  young  learners,  Compound  Num- 
bers should  be  introduced  before,  instead  of  after,  Fractions — for  the 
following  reasons: 

1st.  The  operations  of  Addition,  Subtraction,  Multiplication,  and 
Division  of  compound  numbers  are  analogous  to  the  same  operations 
in  simple  numbe/s,  and  serve  to  illustrate  the  principles  of  the  Fun- 
damental Rules. 

2d.  The  subject  of  Fractions  is  important  and  difficult.  Before 
studying  it,  most  pupils  require  more  mental  discipline  than  is 
furnished  by  the  elementary  rules;  this  is  acquired  by  the  study  of 
Compound  Numbers. 

3d.  The  general  principles  involved  in  their  study,  do  not  require 
a  knowledge  of  Fractions.  The  examples  involving  fractions  are 
few,  and  are  introduced,  as  they  should  be,  with  other  exercises  in 
that  subject. 

Teachers,  who  prefer  it,  can  direct  their  pupils  to  defer  Compound 
Numbers  until  they  have  studied  Fractions  as  far  as  page  159. 

DEFINITIONS. 

50.  A  compound  number  is  made  up  of  two  or 
more  concrete  numbers  of  different  denominations ;  as,  3 
pecks   7   quarts   1    pint. 


Rem.   1. — The   different   denominations    of  a   compound   number 
must    belong   to   the  same   table;   thus,   in   the   example   given,   the 

(71) 


72  RAY\S  NEW  PRACTICAL  ARITHMETIC. 

pecks  may  be  reduced  to  quarts  or  pints,  and  the  pints  and  quarts 
are  j)arU  of  u  peck.  3  pecks  7  dollars  would  not  be  a  compound 
number. 

Rem.  2. — Compound  numbers  resemble  simple  numbers  in  tho 
following  particulars:  the  denominations  of  compound  numbers  cor- 
respond to  the  orders  of  simple  numbers,  and  a  certain  number  of 
units  of  a  lower  denomination  make  one  unit  of  the  next  higher 
denomination. 

Rem.  -5. —  Most  compound  numbers  differ  from  simple  numbers 
in  this;  ten  units  of  each  lower  denomination  do  not  uniformly 
make  one  unit  of  the  next  higher  denomination. 

Rem.  4. — In  United  States  Money  and  the  Metric  System  of 
Weights  and  Measures,  however,  ten  units  of  a  lower  denomination 
do  make  one  imit  of  the  next  higher  denomination. 

51.  1.  The  operations  with  compound  numbers  are 
Bediictlon,  Addition,  Subtraction,  Multiplication,  and  Di- 
vision. 

2.  Reduction  is  the  process  of  changing  the  denom- 
ination of  a  number  without  aUering  its  value. 

Thus,  5  yards  may  be  changed  to  feet;  for,  in  1  yard  there  are  3 
feet;  then,  in  5  yards  there  are  5  times  3  feet,  which  are  15  feet. 

3.  Eeduction  takes  place  in  two  ways:  1st.  From  a 
higher  denomination  to  a  lower.  2d.  From  a  lower  de- 
nomination  to  a  higher. 

UNITED   STATES  MONEY. 

5"2.  United  States  money  is  the  money  of  the  United 
States  of  America. 

Table. 

10  mills,  m.,  make  1  cent,  marked  ct. 

10  cents                  "     1  dime,             "  d. 

10  dimes                 ^'     1  dollar,           "  %. 

10  dollars                "     1  eagle,             "  E. 


UNITED  STATES  MONEY. 


73 


Rem.  1. — United  States  money  was  established,  by  act  of  Con- 
gress, in  1786.  The  first  money  coined,  by  the  authority  of  the 
United  States,  was  in  1793.  The  coins  first  made  were  copper 
cents.  In  1794  silver  dollars  were  made.  Gold  eagles  were  made 
in  1795;  gold  dollars,  in  1849.  Gold  and  silver  are  now  both  legally 
standard.     The  trade  dollar  was  minted  for  Asiatic  commerce. 

Rem.  2. — The  coins  of  the  United  States  are  classed  as  bronze, 
nickel,  silver,  and  gold.  The  name,  value,  composition,  and  weight 
of  each  coin  are  shown  in  the  following 


Table. 


r                       1 

COIN.            VALUE. 

COMPOSITION. 

WEIGHT. 

48      grains  Troy. 

BRONZK.         1 

One  cent.           !    1  cent. 

95  parts  copper,  5  parts  tin  &  zinc. 

NICKEL. 

3-cent  piece. 
5-cent  piece. 

3  cents. 
5  cents. 

75  parts  copper,  25  parts  nickel. 
75     "           *'       25     •' 

30      grains  Troy. 
73.1G      " 

SILVER. 

Dime. 

Quarter  dollar. 
Half  dollar. 
Dollar. 

10  cents. 
25  cents. 
50  cents. 
100  cents. 

90  parts  silver,  10  parts  copper. 
90     "         "       10      " 
90     "         "       10      " 
90     '*         "        10      *' 

2.5  grams. 
6.25      " 
12.5 
4123^  grains  Troy. 

GOLD. 

Dollar. 

Quarter  eagle. 
Three  dollar. 
Half  eagle. 
Eagle. 
Double  eagle. 

100  cents. 
23^  dollars. 
3  dollars. 
5  dollars. 
10  dollars. 
20  dollars. 

90  parts  gold.  10  parts  copper. 

90      "        "       10      " 

90      "        "       10      " 

90      •'        "       10      " 

90      "        "       10     " 

90      "        "       10      " 

25.8  grains  Troy. 

64.5 

77.4 
129 
258 
516 

Rem.  3.  — a  deviation  in  weight  of  |  a  grain  to  each  piece,  is 
allowed  by  law  in  the  coinage  of  Double  Eagles  and  Eagles;  of  \  of 
a  grain  in  Half  Eagles  and  the  other  gold  pieces:  of  1^  grains  in  all 


74  KAY'S    NEW   PRACTICAL   ARITHMETIC. 

silver  pieces;  of  3  grains  in  the  iivc-cent  piece;   and  of  2  grains  in 
the  three-cent  piece  and  one  cent. 

Rem.  4. — The  mill  is  not  coined.     It  is  used  only  in  calculations. 

53.  1.  A  sum  of  money  is  expressed  as  dollars  and 
cents,  and,  when  written  in  figures,  is  always  preceded 
by  the  dollar  sign  ($). 

Rem. — Calculations  are  sometimes  carried  out  to  mills,  but,  in  bus- 
iness transactions,  the  final'result  is  always  taken  to  the  nearest  cent. 

2.  A  period  (.),  called  the  decimal  point,  is  used  to 
separate  the  dollars  and  cents. 

3.  Eagles  are  read  as  tens  of  dollars,  and  dimes  are 
read  as  tens  of  cents. 

Thus,  $24.56  is  read  24  dollars  56  cents;  not  2  eagles 
4  dollars  5  dimes  6  cents.  $16,375  is  read  16  dollars 
37  cents  5  mills. 

4.  Hence,  the  figures  to  the  left  of  the  decimal  point  ex- 
press a,  number  of  dollars;  the  two  figures  to  the  right  of 
the  decimal  point,  a  number  of  cents;  and  the  third,  figure 
to  the  right,  mills. 

Rem. — If  the  number  of  cents  is  less  than  10,  a  cipher  must  be  put 
in  the  tens'  place. 

EXAMPT^ES   TO  BE  ^VRITTEX. 

1.  Twelve  dollars  seventeen  cents  eight  mills. 

2.  Six  dollars  six  cents  six  mills.  ^ 

3.  Seven  dollars  seven  ,  mills. 

4.  Forty  dollars  fifty -three  cents  five  mills. 

5.  Two  dollars  three  cents. 

6.  Twenty  dollars  two  cents  two  mills. 

7.  One  hundred  dollars  ten  cents. 

8.  Two  hundred  dollars  two  cents. 

9.  Four  hundred  dollars  one  cent  eicfht  mills. 


UNITED  STATES  MONEY.  75 

EXAMPLES    TO   BE    READ. 

$18,625  $  70.015  ^6.12  $   29.00 

320.324  $100.28  $3.06  $100.03 

$79.05  $150.05  $4.31  $  20.05 

$46.00  $100.00  $5.43  $  40.125 

REDUCTION    OF    IT.    S.    MONEY. 

54.  1.  As  there  are  10  mills  in  1  cent,  in  any  num- 
ber of  cents  there  are  10  times  as  many  mills  as  cents 
Therefore  to  reduce  cents  to  mills — 

Rule. — Multiply  the  number  of  cents  by  ten ;  that  is^ 
annex  one  cipher. 

2.  Conversely,  to  reduce  mills  to  cents — 

Rule. — Divide  the  iiumber  of  mills  by  ten ;  thcit  is^  cut  off 
one  figure  from  the  right. 

3.  As  there  are  10  cents  in  1  dime  and  10  dimes  in 
1  dollar,  there  are  10  X  10  =  100  cents  in  1  dollar;  then, 
in  any  number  of  dollars  there  are  100  times  as  many 
cents  as  dollars.     Therefore,  to  reduce  dollars  to  cents — 

Rule. — Multiply  the  number  of  dollars  by  one  hundred; 
that  is,  annex  two  ciphers. 

4.  Conversely,  to  reduce  cents  to  dollars-^ 

Rule. — Divide  the  number  of  cents  by  07ie  hundred ;  that 
is,  cut  off  tiro  figures  from  the  right. 

5.  As  there  are  10  mills  in  1  cent  and  100  cents  in  1 
dollar,  there  are  100  X  lOirrrlOOO  mills  in  1  dollar;  then, 
in  any  number  of  dollars  there  are  1000  times  as  many 
mills  as  dollars.     Therefore,  to  reduce   dollars  to  mitls — 


76 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Rule. — Multiply  the  number  of  dollars  by  one  thousand; 
that  is,  annex  three  ciphers. 

6.  Conversely,  to  reduce  mills  to  dollars. 

Rule. — Divide  the  number  of  mills  by  one  thousand :  that 
is,  cut  off  three  figures  from  the  right. 

55.  The  reduction  of  mills  or  cents  to  dollars  may  be 
made  simply  with  the  decimal  point.     Thus, 

1st.  If  the  sum  is  mills.  Rule. — Put  the  decimal  point 
between  the  third  and  fourth  figures  from  the  right. 

2d.  If  the  sum  is  cents.  Rule. — Put  the  decimal  point 
betiveen  the  second  and  third  figures  from  the  right. 


1.  Eeduce  17  ct.  to  mills.  170  m 

2.  Ileduce  28  ct.  to  mills.  280  m 

3.  Keduce  43  ct.  and  6  m.  to  mills.  436  m 

4.  Reduce  70  ct.  and  6  m.  to  mills.  706  m. 

5.  Reduce  106  m.  to  cents.  10  ct.  6  m. 

6.  Reduce  490  mills  to  cents.  49  ct. 

7.  Reduce  9  dollars  to  cents.  900  ct. 

8.  Reduce  14  dollars  to  cents.  1400  ct. 

9.  Reduce  104  dollars  to  cents.  10400  ct. 

10.  Reduce  $60  and  13  ct.  to  cents.  6013  ct. 

11.  Reduce  $40  and  5  ct.  to  cents.  4005  ct. 

12.  Reduce  375  ct.  to  dollars.  83.75 

13.  Reduce  9004  ct.  to  dollars.  S90.04 

14.  Reduce  4  dollars  to  mills.  4000  m. 

15.  Reduce  S14  and'*2  ct.  to  mills.  14020  m. 

16.  Reduce  2465  mills  to  dollars.  S2.46  5. 

17.  Reduce  3007  mills  to  dollars.  $3.00  7. 

18.  Reduce  3187  cents  to  dollars.  $31.87. 

19.  Reduce  10375  mills  to  dollars.  $10,375. 


UNITED  STATES  MONEY.  77 


ADDITION    OF    U.    S.    MONEY. 

56.  1.  Add  together  4  dollars  12  cents  5  mills;  7 
dollars  6  cents  2  mills;  20  dollars  43  cents;  10  dollars 
5  mills;  16  dollars  87  cents  5  mills. 

Rule. — 1.  Write  the  numbers  and  add  as  orERATiox. 
in  simple   numbers.  $•  ct.  m. 

2.  Place  the  decimal  point  in  the  sum  under  7  *  o  fi  9 

the  decimal  points  above.  2  0 !  4  3  0 

10.005 

Proof. — The  same  as  in  Addition  of  Sim-         16.875 

pie  Numbers.  $58749T 

2.  What  is  the  sum  of  17  dollars  15  cents ;  23  dollars 
43  cents;  7  dollars  19  cents;  8  dollars  37  cents;  and  12 
dollars  31  cents?  S68.45. 

3.  Add  18  dollars  4  cents  1  mill;  16  dollars  31  cents 
7  mills;  100  dollars  50  cents  3  mills;  and  87  dollars  33 
cents  8  mills.  S222.199. 

4.  William  had  the  following  bills  for  collection:  $43.75; 
$29.18;  $17.63;  $268.95;  and  $718.07:  how  much  was  to 
be  collected?  $1077.58. 

5.  Bought  a  gig  for  $200 ;  a  watch  for  $43.87 ;  a  suit 
of  clothes  for  $56.93 ;  a  hat  for  $8.50 ;  and  a  whip  for 
$2.31:  w^hat  was  the  amount?  $311.61. 

6.  A  person  has  due  him,  five  hundred  and  four  dol- 
lars six  cents;  $420.19;  one  hundred  and  ^yq  dollars 
fifty  cents ;  $304 ;  $888.47 :  what  is  the  whole  amount 
due  him?  $2222.22. 

7.  Add  five  dollars  seven  cents;  thirty  dollars  twenty 
cents  three  mills ;  one  hundred  dollars  five  mills ;  sixty 
dollars  two  cents;  seven  hundred  dollars  one  cent  one 
mill;  $1000.10;   forty  dollars  four  mills;  and  $64.58  7. 

$2000. 


78  HAY'S  NEW  PRACTICAL  ARITHMETIC. 

SUBTRACTION    OF   U.    S.    MONEY. 

57.     1     From    one    hundred    dollars    five    cents    threo 
mills,  tuke    eighty  dollars  twenty  cents  and  seven  milla 

Rule. — 1.    Write   the  numbers  and  subtract 
as   in    Simple  Numbers.  operatiox. 

2.  Flace  the  decimal  point  in  the  remainder  $.    et.  m. 

under  the  decimal  points  above.  100.053 

80.207 


Proof. — The   same   as   in    Subtraction   of       $19,846 
Simple   Numbers. 

2.  From  $29,342  take  817.265.  $12,077. 

3.  From  $46.28     take  $17.75.  $28.53. 

4.  From  $20.05     take  $5.50.  $14.55. 

5.  From  $3,  take  3  et.  $2.97. 

6.  From  $10,  take  1  mill.  $9,999. 

7.  From  $50,  take  50  ct.  5  mills.  $49,495. 

8.  From  one  thousand  dollars,  take  one  dollar  one 
cent  and  one  mill.  $998,989. 

9.  B  owes  1000  dollars  43  cents;  if  he  pay  nine  hun- 
dred dollars  sixty-eight  cents,  how  much  =^^%e  still 
owe?  ""        $99.75. 

MULTIPLICATION   OF   U.    S.    MONEY. 

58.  1.  What  will  13  cows  cost,  at  47  dollars  12  cents 
5  mills  each? 

Rule. — 1.  Multiply  as  in  Simple  Numbers.       operation. 
2.   Put  the  decimal  point  in  the  same  place  $47,125 

in  the  product^  as   it   is   in  the  multiplicand. 


1_3 

141375 
47125 


Proof. — The  same  as  in  Multiplication  of 
Simple  Numbers.  $612,6  2  5 


UNITED  STATES  MONEY.  79 

2.  Multiply  S7.835  by  8.  $62.68 

3.  Multiply  $12,  9  et.  3  m.  by  9.  $108,837, 

4.  Multiply  $23,  1  ct.  8  m.  by  16.  $368,288 

5.  Multiply  $35,  14  ct.  by  53.  $1862.42, 

6.  Multiply  $125,  2  ct.  by  62.  $7751.24. 

7.  Multiply  $40,  4  ct.  by  102.  $4084.08, 

8.  Multiply  12  ct.  5  m.  by  17.  $2,125, 

9.  Multiply  $3.28  by  38.  $124.64 

10.  What  cost  338  barrels  of  cider,  at  1  dollar  6  cents 
a  barrel?  $358.2B. 

11.  Sold    38    cords   of  wood,  at    5    dollars    75   cents  a 
cord:  to  what  did  it  amount?  $218.50. 

12.  At  7  ct.  a  pound,  what  cost  465  pounds  of  sugar  ^ 

Note. — Instead  of  multiplying  7  cents  l)y  465,  mul-  operation 

tiply  465  by  7,  which  gives  the  same  product,  Art.  30.  4  6 1 

But,  to  place  the  decimal  point,  remember  that  7  cents  .0  ? 

is  the  true  multiplicand.  $  'S^2.6 1 

13.  What  cost  89  yards  of  sheeting,  at  34  ct.  a  yard? 

$30.26. 

14.  What  will  24  yards  of  cloth  cost,  at  $5.67  a  yard? 

$136.08. 

15.  I    have    169  sheep,  valued  at    $2.69  each  :    what  is 
the  value  of  the  whole?  $454.61. 

16.  If  I  sell   691   bushels  of  wheat,  at  $1.25  a  bushel, 
what  will  it  amount  to?  $863.75. 

17.  I    sold    73    hogsheads    of  molasses,    of  63    gallons 
each,  at  55  ct.  a  gallon:  what  is  the  sum?  $2529.45. 

18.  What  cost  4  barrels  of  sugar,  of  281  pounds  each, 
at  6  cents  5  mills  a  pound?  $73.06. 

19.  Bought    35  bolts  of  tape,  of  10    yards    each,  at  1 
cent  a  yard  :  what  did  it  cost  ?  $3.50. 

20.  If  I  earn  13  ct.  an  hour,  and  work  11  hours  a  day, 
how  much  will  I  earn  in  312  days?  $446.16. 


80  KAY'S  NEW  PEACTICAL  AKITHMETIC. 

21.  I   sold    18   bags    of  wheat,  of  3  bushels   each,    at 
$1.25  a  bushel:  what  is  the  amount?  $67.50. 

22.  What    cost  150  acres  of  land,  at  10  dollars  1  mill 
per  acre?  81500.15. 

23.  What  cost  17  bags  of  coffee,  of  51  pounds  each,  at 
24  cents  7  mills  per  pound?  $214.14  9. 

DIVISION    OF    U.    S.    MONEY. 

59.     Case  I. — To  find    how   many  times  one  sum    of 
money  is  contained  in  another. 

1.  How  much  cloth,  at  7  cents  a  yard,  will  $1.75  buy? 

OPEllATIQN. 

Solution.— Aij  many  yards  as  7  cents  is  contained  7)175 

times  in  175  cents,  which  are  25.  2  5 

Rule. — 1.  Beduce  both  sums  of  money  to  the  same  denom- 
ination. 

2.  Divide  as  in  Simple  Numbers. 

2.  How  much  rice,  at  9  cents  a  pound,  can  be  bought 
for  72  cents?  8  lb. 

3.  How  many  towels,  at   37    cents  and  5  mills  apiece, 
can  be  bought  for  $6?  16. 

4.  How  many  yards   of  calico,  at  8  cents  a  3^ard,  can 
be  bought  for  $2.80?  35  yd. 

5.  How   many  yards    of  ribbon,  at    25   cents    a    yard, 
can  be  purchased  for  $3?  12  yd. 

6.  At   $8.05  a  barrel,  how    many  barrels  of  flour  will 
$161  purchase?  20  bl. 

7.  At  7  cents  5  mills  each,  how  many  oranges  can  be 
bought  for  $1.20?  "  16. 

8.  At  $1,125  per  bushel,  how  many  bushels  of  wheat 
can  be  purchased  for  $234?  208  bu. 


UNITED  STATES  MONEY.  81 

Case  II. — To  divide  a  sum  of  money  into  a  given 
number  of  equal  parts. 

1,  A  man  worked  3  days  for  $3.75,  what  were  his 
daily  wages? 

OPERATION 

Solution. — His  daily  wages  were  $3.75  -^  3  r=:  $1.25.         3  )  3.7  5 

$T25 

2.  A  farmer  sold  6  bushels  of  wheat  for  $d :  how  much 
a  bushel  did  he  get? 

Solution. — He  got  for  eadi  bushel  $9—6.    $9  di-  operation. 

vided   by   6   gives   a  quotient   $1,    with   a   remainder  6)9.00 

$3  =  300  cents.     300  cents  divided  by  6  gives  a  quo-  $1.50 
tient  50  cents. 

Rule. — 1.  Divide  as  in  Simple  Nvmbers, 

2.  Put  the  decimal  point  in  the  same  j^lace  in  the  quotient 
as  it  is  in,  the  dividend, 

Kem.  1.— If  the  dividend  is  dollars,  and  the  division  not  exact, 
annex  two  ciphers  after  the  decimal  point  for  cents;  and,  if  nec- 
essary, a  third  cipher  for  mills. 

Rem.  2. — Should  there  be  a  remainder  after  obtaining  tlie  mills, 
it  may  be  indicated  by  the  sign  -f  placed  after  the  quotient. 

3.  Divide  65  dollars  equally  among  8  persons.     S8.125. 

4.  A  farmer  received  $29.61  for  23  bushels  of  wheat: 
how  much  was  that  per  bushel?  $1,287  H-. 

5.  If  4  acres  of  land  cost  $92.25,  how  much  is  that  an 
acre?  $23,062+. 

6.  Make  an  equal  division  of  $57.50  among  8  persons. 

$7,187+. 

7.  A  man  received  $25.76  for  16  days'  work:  how 
much  was  that  a  day?  $1,61. 


82  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

8.  I  bought  755  bushels  of  apples  for  $328,425:  what 
did  they  cost  a  bushel?  $0,435. 

9.  My  salary  is  $800  a  year :  how  much  is  that  a  day, 
there  being  313  working  days  in  the  year?         $2.555-|-. 

10.  Divide  ten  thousand  dollars  equally  among  133 
men:  what  is  each  man's  share?  $75.187 -f. 

11.  A  man  purchased  a  farm  of  154  acres,  for  two 
thousand  seven  hundred  and  five  dollars  and  1  cent: 
what  did  it  cost  per  acre?  $17,565. 

12.  I  sold  15  kegs  of  butter,  of  25  pounds  each,  for 
$60:  how  much  Avas  that  a  pound?  16  ct. 

13.  I  bought  8  barrels  of  sugar,  of  235  pounds  each, 
for  $122.20:  what  did   1   pound  cost?  $0,065. 


Promiscuous  Examples. 

60.  1.  I  oAve  A  $47.50;  B,  $38.45;  C,  $15.47;  D, 
$19.43:  what  sum  do  I  owe?  $120.85. 

2.  A  owes  $35.25 ;  B,  $23.75 ;  C,  as  much  as  A  and  B, 
and  $1  more:  what  is  the  amount?  $119. 

il  A  paid  me  $18.38;  B,  $81.62;  C,  twice  as  much  as 
A  and  B:  how  much  did  I  receive?  $300. 

4.  I  went  to  market  w4th  $5 ;  I  spent  for  butter  75 
cents,  for  eggs  35  cents,  for  vegetables  50  cents,  for  flour 
$1.50:  how  much  money  was  left?  $1.90. 

5.  A  lady  had  $20;  she  bought  a  dress  for  $8.10,  shoes 
for  $5.65,  eight  yards  of  delaine  at  25  cents  a  yard,  and 
a  shawl  for  $4:  what  sum  was  left?  25  ct. 

6.  I  get  $50  a  month,  and  spend  $30.50  of  it:  how 
much  will  I  have  left  in  6  months?  $117. 

7.  A  farmer  sold  his  marketing  for  $21.75  :  he  paid  for 
sugar  $3.85,  for  tea  $1.25,  for  coffee  $2.50,  for  spices 
SI .50:  how  much  had  he  left?  $12.65. 


UNITED  STATES  MONEY.  83 

8.  T  owe  A  S37.06;  B,  $200.85;  C,  S400;  D,  $236.75, 
and  E  $124.34;  my  property  is  worth  $889.25  :  how  much 
do  I  owe  more  than  I  am  worth?  $109.75. 

9.  Bought  143  pounds  of  coffee,  at  23  cents  a  pound : 
after  paying  $12.60,  what  was  due?  $20.29. 

10.  A  owed  me  $400:  he  paid  me  435  bushels  of  corn, 
at  45  cents  a  bushel:  what  sum  is  due?  $204.25. 

11.  If  B  spend  65  cents  a  day,  how  much  will  he  save 
in  365  days,  his  income  being  $400?  $162.75. 

12.  Bought  21  barrels  of  apples,  of  3  bushels  each,  at 
35  cents  a  bushel:  what  did  they  cost?  $22.05. 

13.  What  cost  four  pieces  of  cambric,  each  containing 
19  yards,  at  23  cents  a  yard?  $17.48. 

14.  If  25  men  perform  a  piece  of  Avork  for  $2000,  and 
spend,  while  doing  it,  $163.75,  what  will  be  each  man's 
share  of  the  profits?  $73.45. 

15.  If  16  men  receive  $516  for  43  days'  work,  how 
much  does  each  man  earn  a  day?  75  ct. 

16.  C  earned  $90  in  40  days,  working  10  hours  a  day: 
how  much  did  he  earn  an  hour?  22  ct.  5  m. 

17.  A  merchant  failing,  has  goods  worth  $1000,  and 
$500  in  cash,  to  be  equally  divided  among  22  creditors: 
how  much  will  each  receive?  $68.18-|-. 


MERCHANTS'    BILL8. 

A  Bill  or  Account,  is  a  written   statement  of  articles 
bought  or  sold,  with  their  prices,  and  entire  cost. 

18.  Bought    9  pounds  Coffee,  at  $0.32  per  lb.     $ 

4  pounds  Tea,        "  1.25     do. 

45  pounds  Sugar     "  .09     do. 

17  pounds  Cheese   "  .20     do. 


"What  is  the  amount  of  my  bill?  $15.33 


84  RAY'S  NEW  PRACTICAL  AKITHMETIC. 

19.  Bought  22  yards  Silk,  at  $1.75  per  yd.  S 

18  yards  Muslin,       ''        .15      do. 
25  yards  Linen,        '-        .65      do. 
6  yards  Gingham,  "        .18      do. 


What  is  the  whole  amount? 

S58.53 

20.  Bought— 

4  pounds  Prunes,                 at  SO.  18 

per  lb. 

$    . 

8  pounds  Peaches,                '' 

.23 

do. 

7  pounds  Rice,                      *' 

.11 

do. 

6  pounds  Oat-meal,             " 

.09 

do. 

13  pounds  Java  Coffee,         " 

.35 

do. 

26  pounds  Sugar,                  " 

.12 

do. 

What  is  the  whole  amount? 

$11.54 

21.  Bought  43  yards  Muslin,     at  $0.13  per  yd.   $ 
28>^5iiaTd8  Calico,       '-       .09       do. 
23  yards  Alpaca,     ''       .23       do. 


What  is  the  whole  amount?  $13.40 

REDUCTION  OF  COMPOUND  NUMBERS. 
DRY   MEASURE. 

61.  Dry  Measure  is  used  in  measuring  grain,  vege- 
tables, fruit,  coal,  etc. 

Table. 

2  pints  (pt.)  make  1  quart,  marked  qt. 
8  quarts  "       1  peck,  ''         pk. 

4  pecks  ''       1  bushel,       '•         bu. 

Rem.  1. — The  standa7'd  iinii  of  Dry  Measure  is  the  bushel;  it  is  a 
cylindrical  measure  18^  inches  in  diameter,  8  inches  deep,  and  con- 
tains 21501  cubic  inches. 


REDUCTION  OF  COMPOUND  NUMBERS. 


85 


Rem.  2.  —When  articles  usually  measured  by  the  above  table  are 
sold  by  weight,  the  bushel  is  taken  as  the  unit.  The  following  table 
gives  the  legal  weight  of  a  bushel  of  various  articles  in  avoirdupois 
pounds: 


ARTICLES. 

LB. 

-  ■■   -■-- 

EXCEPTIONS. 

Beans. 

60 

Blue  Grass  Seed. 

14 

Clover  Seed. 

60 

N.  J.,  64. 

Coal  (mineral). 

80 

Ind.,  70. 

Corn  (shelled). 
Flax  Seed. 

56 
56 

Cal.,  111.   Mo.,  52. 
111.,  N.  J.,  N.  Y.,  55. 

Hemp  Seed. 
Oats. 

44 
32 

rConn.,  28;   Me.,  N.  H.,   Mass.,  N.  J., 

30; 

(Ky.,    33^;    Oregon,    34;    la.,  32 ;  Mo, 

35. 

Potatoes. 

60 

Rye. 

56 

La.,  52;  Cal.,  111.,  54. 

Salt.     ■ 

50 

N.  Y.,  56. 

Timothy  Seed. 

45 

N.  Y.,  44;  Wis.,  46. 

Wheat. 

60 

Conn.,  56. 

To  Teachers. — Numerous  questions  should  be  asked  on  each  table 
similar  to  the  following: 


1.  How  many  pints  in  2  quarts?      In  4?      In  6?     In 
8?     In  10? 

2.  How  many  quarts  in  3  pk.?     In  5?     In  7?     In  9? 

3.  How  many  pecks    in    9    bii.?     In  11?     In  13?     In 
15?     In  17?     In  19? 

4.  How  many  quarts  in  10  bu.?     In  12?     In  14?     In 
18?     In  25?     In  56? 

5.  How  many  pecks  in  16  qt.?      In  24?      In    32?     In 
40?     In  48?     In  64? 

6.  How  many  bushels  in  32  qt.?     In  64?     In  96? 

7.  How  many  pints  in  1  bu.  ?     In  2  ?     In  5  ? 


St)  KAY  S  NEW  PRACTICAL  ARITHMETIC. 

62.  The  preceding  examples  show  that — 

To  reduce  quarts  to  pints,  multiply  the  number  of 
quarts  by  the  number  of  pints  in  a  quart. 

To  reduce  pecks  to  quarts,  or  bushels  to  pocks,  multi- 
ply in  the  same  manner. 

Hence,  to  reduce  from  a  higher  to  a  lower  denomina- 
tion, multiply  by  the  number  of  units  that  make  one  unit 
of  the  required  denomination. 

They  also  show  that — 

To  reduce  pints  to  quarts,  divide  the  number  of  pints 
by  the  number  of  pints  in  a  quart. 

To  reduce  quarts  to  pecks,  or  pecks  to  bushels,  divide 
in  the  same  manner. 

Hence,  to  reduce  from  a  lower  to  a  higher  denomina- 
tion, divide  by  the  number  of  units  that  make  one  unit 
of  the  required  higher  denomination. 

1.  Eeduce  3  bushels  to  pints. 

OPERATION. 

Solution. — To  reduce  bu.  to  pk.  multiply  by  4,  ^  bu. 
because  there  are  4  pk.  in   1   bu.,  or  4  times  as 
many  pk.  as  bu.     To  reduce  pk,  to  qt.  multiply  by 

8,  because  there  are  8  qt.  in  1  pk.     To  reduce  qt.  ^    . 

to  pt.  multiply  by  2,  because  there  are  2  pt.  in  2     * 

1  qt.  foTpt. 

2.  Eeduce  192  pints  to  bushels. 

OPERATION. 

Solution. — To  reduce  pt.  to  qt.  divide  by  2,  be-  2)192  pt. 

cause  there  are  2  pt.  in  1  qt.     To  reduce  qt.  to  pk.  8)96  qt. 

divide  by  8,  because  there  are  8  qt.  in  1  pk.     To  4)12  pk. 

reduce  pk.  to  bu.  divide  by  4,  because  there  are  4  3  bu. 
pk.  in  1  bu. 

The  two  preceding  examples  show  that  reduction  from 
a  higher  to  a  lower  denomination^  and  from  a  loicer  to  a 
higher  denomination,  prove  each  other. 


f2pk. 


REDUCTION  OF  COMPOUND  NUMBERS.  87 

3.  Eeduco  7  bu.  3  j)k.  6  qt.  1  pt.  to  joints. 


OPERATION. 

bu.  pk.  qt.  pt. 
Solution.  —  Multiply    the    bu.    by   4,  7      3     6     1 

making  28  pk.,  and  add  the  3  pk.     Then         _4 
multiply  the  31  pk.  by  8  and  add  the  6        3  1  pk.  in  7  bu.  3  pk 

qt.;  multiply  the  254  qt.  by  2  and  add  the     z. 

1  pt.;  the  result  is  509  pt.  2  5  4  qt.  in  31  pk.  6  qt. 


2 

5  0  9  pt.  in  the  whole. 


4.  Eediice  509  pt.  to  bushels. 


Solution. — To  reduce  pt.  to  qt.  divide  by  operation. 

2,  and   there  is  1  left;    as  the  dividend  is  pt.  2)509 

the  remainder  must  be  pt.     To  reduce  qt.  to  8)254  qt.  1  pt. 

pk.  divide  by  8,  and  6  qt.  are  left.     To  reduce  4)31  pk.  6  qt. 

pk.  to  bu.  divide  by  4,  and  3  pk.  are  left.     The  7  bu.  3  pk. 
answer  is,  therefore,  7  bu.  3  pk.  6  qt.  1  pt. 


03,     RULES   FOR   REDUCTION. 
I.    FROM    A    HIGHER   TO    A    LOWER    DENOMINATION. 

1.  Multiply  the  highest  denomination  given,  by  that  num 
her  of  the  next  lower  which  makes  a  unit  of  the  higher. 

2.  Add  to  the  product  the  number,  if  any,  of  the  lower 
denomination. 

3.  Proceed  in  like  manner  icith  the  result  thus  obtained, 
tdl  the  u'hole  is  reduced  to  the  required  denomination. 

II.     FROM    A    LOWER    TO    A    HIGHER     DENOMINATION. 

1.  Divide   the  given  quantity  by  that  number  of  its  own 
denomination  which  makes  a  unit  of  the  next  higher. 

2.  Proceed  in  like  manner  with  the  quotient  thus  obtained, 
till  the  whole  is  reduced  to  the  required  denomination. 


88  KAY'S  NEW  PKACTlCAl^  AKITHMETIC. 

3.  The  last  quotient,  with  the  several  remainders^  if  any, 
annexed,  will  be  the  answer. 

Proof. — Eevcrsc  I  he  operation:  that  is,  reduce  the 
answer  to  the  denomination  from  which  it  was  derived. 
Jf  this  result  is  the  same  as  the  quantity  given,  the  work 
is  correct. 

5.  lieduce  4  bu.  2  pk.  1  qt.  to  pints.  290  pt. 

6.  Eeduce  7  bu.  3  pk.  7  qt.  1  pint  to  pints.  511  pt. 

7.  Reduce  3  bu.  1  pt.  to  pints.  193  pt. 

8.  Reduce  384  pt.  to  bushels.  (>  bu. 

9.  Reduce    47  pt.  to  pecks.                  2  pk.  7  qt.  1  pt. 

10.  Reduce     95  pt.  to  bushels.       1  bu.  1  pk.  7  qt.  1  pt. 

11.  Reduce  508  pt.  to  bushels.  7  bu.  3  pk.  G  qt. 

LIQUID  MEASURE. 
64.     Liquid  Measure  is  used  for  measuring  all  liquids. 

Table. 

4  gills  (gi.)  make  1  pint,  marked  pt. 
2  pints  ^*       1  quart,      '•         qt. 

4  quarts  "       1  gallon,    "         gal. 

Kem. — The  standard  unit  of  liquid  measure  is  the  gallon,  which 
contains  231  cubic  inches. 

1.  Reduce       17  gal.  to  pints.  136  pt 

2.  Reduce       13  gal.  to  gills.  416  gi. 

3.  Reduce     126  gal.  to  pints.  1008  pt 

4.  Reduce  1260  gal.  to  gills.  40320  gi. 

5.  Reduce  1120  gi.     to  gallons.  35  gal. 

6.  How  many  gallons  in  1848  cubic  inches?  8  gal. 

7.  How  many  gallons  in  a  vessel  containing  138138 
cubic  inches?  598  gal. 


REDUCTION  OF  COMPOUND  NUMBEKS.  89 


AVOIRDUPOIS  WEIGHT. 

65.  Avoirdupois  Weight  is  used  for  weighing  all 
ordinary  articles. 

Table. 

16  ounces  (oz.)      make    1  pound,  ''        lb. 

100  pounds  "       1  hundred-weight/'        cwt. 

20  cwt.,  or  2000  lb.,"       1  ton,  ^  "        T. 

Rem.  1. — The  standard  avoirdupois  pound  of  the  United  States 
is  determined  from  the  Troy  pound,  and  contains  7000  grains  Troy. 

Rem.  2. — At  the  Custom  House  (and  in  some  trades)  2240  pounds 
are  considered  a  ton. 

1.  Reduce  2  cwt.  to  pounds.  200. 

2.  Eeduce  3  cwt.  75  lb.  to  pounds.  375. 

3.  Keduce  1  T.  2  cwt.  to  pounds.  2200. 

4.  Eeduce  3  T.  75  lb.  to  pounds.  6075. 

5.  Eeduce  4  cwt.  44  lb.  to  pounds.  444. 

6.  Eeduce  5  T.  90  lb.  to  pounds.  10090. 

7.  Eeduce  2  cwt.  77  lb.  12  oz.  to  ounces.  4444. 

8.  Eeduce  2  cwt.  17  lb.  3  oz.  to  ounces.  3475. 

9.  Eeduce  1  T.  6  cwt.  4  lb.  2  oz.  to  ounces.  41666. 

10.  Eeduce  4803  lb.  to  cwt.  48  cwt.  3  lb. 

11.  Eeduce  22400  lb.  to  tons.  11  T.  4  cwt. 

12.  Eeduce  2048000  oz.  to  tons.  64  T. 

13.  Eeduce  64546  oz.  to  cwt.  40  cwt.  34  lb.  2  oz. 

14.  Eeduce  97203  oz.  to  tons.  3  T.  75  lb.  3  oz. 

15.  Eeduce  544272  oz.  to  tons.  17  T.  17  lb. 

16.  What  is  the  total  weight  of  52  parcels,  each  con- 
taining 18  lb.?  9  cwt.  36  lb. 

17.  What  is  the  weight  of  180  iron  castings,  each 
weiffhin^   75    lb.?  6  T.  15  cwt. 


90  KAY'S  NEW  PRACTICAL  A  KIT  II  MET  I C. 


LONG    MEASURE. 

66.  Long  Measure  is  used  in  measuring  distances, 
or  lengtli,  in  any  direction. 

Table. 

12  inches  (in.)  make  1  foot,    marked  ft. 

3  feet  '•  1  yard,        '-         yd. 

5^  yards,  or  16^  feet,       '*  1  rod,  ''         rd. 

320  rods  "  1  mile,         "         mi. 

Rem. — The  standard  unit  of  length  is  the  yard.  The  standard 
yard  for  the  United  States  is  preserved  at  Washington.  A  <•(>))> 
of  this  standard  is  kept  at  each  state  capital. 

1.  Reduce  2  yd.  2  ft.  7  in.  to  inches.  103  in. 

2.  Reduce  7  yd.  11  in.  to  inches,  263  in. 

3.  Reduce  12  mi.  to  rods.  3840  rd. 

4.  Reduce  7  mi.  240  rd.  to  rods.  2480  rd. 

5.  Reduce  9  mi.  31  rd.  to  rods.  2911  rd. 

6.  Reduce  133  in.  to  yards.  3  yd.  2  ft.  1  in. 

7.  Reduce  181  in.  to  yards.  5  yd.  1  in. 
8  Reduce  2240  rd.  to  miles.  7  mi. 
9.  Reduce  2200  rd.  to  miles.  6  mi.  280  rd. 

10.  Reduce  1  mi.  to  yards.  1760  y(\. 

11.  Reduce  1  mi.  to  feet.  5280  fL 


SQUARE   MEASURE. 

67.     Square  Measure  is  used  in  measuring  any  thing 
which  has  both  length  and  breadth ;   that  is,  two  dimen- 


A  figure  having  4  equal  sides  and  4  right  angles  is  a 
square. 


REDUCTION  OF  COMPOUND  NUMBERS. 


91 


A  square  inch  is  a  square,  each  side  of  which  is  1  inch 
in  length. 

A  square  foot  is  a  square,  each  side  of  which  is  1  foot. 

A  square  yard  is  a  square,  each  side  of  which  is  1 
yard  (3  feet). 


One 

Square 

Foot. 

Suppose  the  figure  to  represent 
a  square  yard.  It  will  then  be  3 
feet  each  way,  and  contain  9 
square  feet.  Each  foot  will  he  12 
inches  each  way,  and  contain  144 
square  inches. 

The  number  of  small  squares  in 
any  large  square  is,  therefore, 
equal  to  the  number  of  units  in 
one  side  multiplied  by  itself. 


Rem. — By  8  feet  square  is  meant  a  square  figure,  each  side  of 
which  is  3  feet,  or  9  square  feet;  but  by  3  square  feet  is  meant  3 
squares  each  one  foot  long  and  one  foot  wide;  therefore,  the  difference 
in  area  between  a  figure  Zfeet  square  and  one  containing  3  square 
feet,  is  6  square  feet. 


Table. 


144     square  inches   make  1  square  foot,  marked  sq.  ft 

Q        finnar.^    fonf  »i  1     Qmior^o    a7«it1  "  en       ^', 


9     square  feet 
30J  square  yards 
'^"     square  rods 
acres 


160 
640 


1  square  yard, 
1  square  rod, 
1  acre, 
1  square  mile, 


1.  Eeduce  8  sq.  yd.  to  square  inches. 

2.  Reduce  4  A.  to  square  rods. 

3.  Eeduce  1  sq.  mi.  to  square  rods. 

4.  Reduce  2  sq.  yd.  3  sq.  ft.  to  sq.  in. 

5.  Eeduce  5  A.  100  sq.  rd.  to  sq.  rd. 

6.  Eeduce  960  sq.  rd.  to  acres. 


sq.  yd. 

sq.  rd. 
A. 

'•         sq.  mi 

10368  sq.  in. 

640  sq.  rd. 

102400  sq.  rd. 

3024  sq.  in. 

900  sq.  rd. 

6  A. 


92  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

7.  Eeduce  3888  sq.  in.  to  square  yards.  3  sq.  yd. 

8.  Eeduce  20000  sq.  rd.  to  acres.  125  A. 

9.  Reduce  515280  sq.  rd.  to  square  miles. 

5  sq.  mi.  20  A.  80  sq.  rd. 
10.  Eeduce  4176  sq.  in.  to  sq.  yd.         3  sq.  yd.  2  sq.  ft. 

68.  A  Rectangle  is  a  figure  having  four  sides  and 
four  right  angles.     See  the  figure  below. 

The  unit  of  measure  for  surfaces,  is  a  square  whose 
side  is  a  linear  unit;  as  a  square  inch,  a  square  foot,  etc. 

The  Area  or  Superficial  contents  of  a  figure,  is  tlie 
number  of  times  it  contains  its  unit  of  measure. 

1.  How  many  square  inches  in  a  board  4  inches  long 
and  3  inches  wide? 

Explanation.— Dividing  each  of  the  longer 
sides  into  4  equal  parts,  the  shorter  sides  into  o 
equal  parts,  and  joining  the  opposite  divisions 
by  straight  lines,  the  surface  is  divided  into 
squares. 

In  each  of  the  longer  rows  there  are  4  squares, 
that  is,  as  many  as  there  are  inches  in  the  longer  side;  and  there  are 
as  many  such  rows  as  there  are  inches  in  the  shorter  side.     Hence, 

The  whole  number  of  squares  in  tbe  board  is  equal  to  the  product 
obtained  by  multiplying  together  the  numbers  representing  the 
length  and  breadth;  that  is,  4  X  3  =  12. 

Rule  for  Finding  the  Area  of  a  Beetangle. — Multiply 
the  length  by  the  breadth ;  the  product  uill  be  the  area. 

Kem. — Both  the  length  and  breadth,  if  not  in  units  of  the  same 
denomination,   should,  be   made   so   before   multiplying. 

2.  In  a  floor  16  feet  long  and  12  feet  wide,  how  many 
square  feet?  192. 


REDUCTION  OF  COMPOUND  NUMBERS.  93 

3.  How  many  square  yards  of  carpeting  will  cover  a 
room  5  yards  long  and  4  yards  wide?  20. 

4.  How  many  square  yards  of  carpeting  will  cover  two 
rooms,  one  18  feet  long  and  12  feet  wide,  the  other  21 
feet  long  and  15  feet,  wide  ?  59. 

5.  How  many  square  yards  in  a  ceiling  18  feet  long 
and  14  feet  wide?  28. 

6.  In  a  field  35  rods  long  and  32  rods  wide,  how^  many 
acres?  7. 

7.  How  much  will  it  cost  to  carpet  two  rooms,  each 
18  feet  long  and  15  feet  wide,  if  the  carpet  costs  $1.25 
per  square  yard?  $75. 

8.  What  will  it  cost  to  plaster  a  ceiling  21  feet  long 
and  18  feet  wide,  at  17  cents  per  square  yard?         S7.14. 

69.  The  Area  of  a  Eectangle  being  equal  to  the  prod- 
uct of  the  length  by  the  breadth,  and  as  the  product 
of  two  numbers,  divided  by  either  of  them,  gives  the 
other  (36,  4);   therefore, 

Rule. — If  the  area  of  a  rectangle  he  divided  by  either  side, 
the  quotient  will  he  the  other  side. 

Illustration. 

In  Example  1,  68,  if  the  area  12  be  divided  by  4, 
the  quotient  3  is  the  width;  or,  divide  12  by  3,  the 
quotient  4  is  the  length. 

Rkm. — Dividing  the  area  of  a  rectangle  by  one  of  its  sides,  is 
really  dividing  the  number  of  squares  in  the  rectangle  "by  the  num- 
ber of  squares  on  one  of  its  sides. 

In  dividing  12  by  4,  the  latter  is  not  4  linear  inches,  but  the  num- 
ber of  square  inches  in  a  rectangle  4  in.  long  and  1  in.  wide.  See 
figure,  Art.  68. 


94 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


1.  A  floor  containing  132  square  feet,  is  11  feet  wide: 
what  is  its  length?  12  il. 

2  A. floor  is  18  feet  long,  and  contains  30  square  yards: 
what  is  its  width?  15  ft. 

3.  A  field  containing  9  acres,  is  45  rods  in  length: 
what  is  its  width?  32  rd. 

4.  A  field  35  rods  wide,  contains  21  acres :  what  is  its 
length?  96  rd. 


SOLID  OR  CUBIC  MEASURE. 

70.  Solid  or  Cubic  Measure  is  used  in  measuring 
things  having  length,  breadth,  and  thickness;  that  is, 
three  dimensions. 

A  Cube  is  a  solid,  having  0  equal  faces,  which  are 
squares. 


Rem. — If  each  side  of  a  cube  is  1  inch 
long,  it  is  called  a  cubic  inch;  if  each  side 
is  3  feet  (1  yard)  long,  as  represented  in 
the  figure,. it  is  a  cubic  or  solid  yard. 

The  base  of  a  cube,  being  1  square 
yard,  contains  3  X  '^  =  9  square  feet;  and 
1  foot  high  on  this  base,  contains  9  solid 
feet;  2  feet  high  contains  9  X  ^  =  18  solid 
feet;  3  feet  high  contains  9X8  =  27  solid 
feet.  Also,  it  may  be  shown  that  1  solid 
or  cubic  foot  contains  12  X  12  X  12  =  1728  solid  or  cubic  inches. 


^.    ^    ^^ 

-,^ 

■r 

■ 

■ 

Hence,  the  number  of  small  cubes  in  any  large  cube,  is 
equal  to  the  length,  breadth,  and  thickness,  multiplied  together. 


Rem. — Any  solid,  whose  corners  resemble  a  cube,  is  a  rectangular 
solid;  boxes  and  cellars  are  generally  of  this  form. 


KEBUCTION  or  COMPOUND  NUMBERS.  95 

The  solid  contents  of  a  rectangular  solid  are  found,  as  in 
the  cube,  by  multiplying  together  the  length,  breadth,  and 
thickness. 

Table. 

1728  cubic  inches  (cu.  in.)  make  1  cubic  foot,  marked  cu.  ft. 

27  cubic  feet  "      1  cubic  j^ard,      ♦'        cu.  yd. 

128  cubic  feet  =  8X4X4  =  8  ft.  long,")^         ,  ^^        ^ 

^^1  cord,  "        C. 

4  ft.  wide,  and  4  ft.  high,  make) 

Hem.  1. — A  cord  foot  is  1  foot  in  length  of  the  pile  which  makes 
a  cord.  It  is  4  feet  wide,  4  feet  high,  and  1  foot  long;  hence,  it 
contains  16  cubic  feet,  and  8  cord  feet  make  1  cord. 

Rem.  2. — A  perch  of  stone  is  a  mass  16J  ft.  long,  1^  ft.  wide,  and 
1  ft.  high,  and  contains  24J  cu.  ft. 

1.  Reduce  2  cu.  yd.  to  cubic  inches.  93312  cu.  in. 

2.  Reduce  28  cords  of  wood  to  cu.  ft.  3584  cu.  ft. 

3.  Reduce  34  cords  of  wood  to  cu.  in.     7520256  cu.  in. 

4.  Reduce  1  cord  of  wood  to  cu.  in.  221184  cu.  in. 

5.  Reduce  63936  cu.  in.  to  cu.  yd.      1  cu.  yd.  10  cu.  ft. 

6.  How  many  cubic  feet  in   a  rectangular  solid,  8  ft. 
'  long,  5  ft.  wide,  4  ft.  thick?  160  cu.  ft. 

7.  How  many  cubic  yards  of  excavation  in  a  cellar  8 
yd.  long,  5  3^d.  wide,  2  yd.  deep?  80  cu.  yd. 

8.  How  many  cubic  yards  in  a  cellar,  18  feet  long,  15 
feet  wide,  7  feet  deep?  70  cu.  3'd. 

9.  In  a  pile  of  wood  40  feet  long,  12  feet  wnde,  and  8 
feet  high,  how  many  cords?  30  C. 

10.  What  will  be  the  cost  of  a  pile  of  wood  80  feet 
long,  8  feet  high,  and  4  feet  thick,  at  S5.50  per  cord? 

$110. 

11.  What  will  be  the  cost  of  excavating  a  cellar  24  ft. 
long,  15  ft.  wide, -and  6  ft.  deep,  at  $1.25  per  cubic  yard 
or  load?  $100. 


96 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


TIME  MEASURE. 


71,     Time  Measure  is  used  in  measuring  time. 

Table. 
60  seconds  (sec.)  make  1  minute,  marked  min. 


60  minutes 

24  hours 
365  days,  6  hours 
100  years 

Also,     7  days 
4  weeks 
12  calendar  months 

365  days 

366  days 


''  1  hour, 

''  1  day, 

'^  1  year, 

^'  1  century 

make  1  week, 


''         hr. 
''         da. 
yr. 
''         cen. 

marked  wk. 
1  month  (nearly),  '^        mon. 
1  year,  ''        yr. 

1  common  year. 
1  leap  year. 


Rem.  1. — The  exact  length  of  the  moan  solar,  or  tropical  year,  is 
365  days,  6  hours,  48  minutes,  46  seconds. 

To  correct  the  error  of  considering  365  days  as  the  length  of  the 
year,  the  following  rule  has  been  adopted: 

Every  year  whose  number  is  not  divisible  by  4  consists  of  365 
days. 

Every  year  whose  number  is  divisible  by  100,  but  not  by  400, 
consists  of  365  days. 

Every  year,  except  the  even  centuries,  whose  number  is  divisible 
by  4,  and  the  even  centuries   divisible  by  400  consist  of  366  days. 

The  year  containing  366  days  is  called  Leap  year,  and  the  extra 
day  is  added  to  February,  giving  it  29,  instead  of  28  days. 

Rem.  2. — Among  nearly  all  civilized  nations  the  year  is  divided 
into  12  calendar  months,  and  numbered,  in  their  order,  as  follows: 


January, 

February, 

March, 

April, 

May, 

June, 


1st  month,  81  days. 

2d         "        28  " 

8d         "        31  " 

4th       "        30  '« 

5th       «'        31  '' 

6th        "        30  " 


July,  7th  month,  31  days. 

August,  8th  **  31  " 
September,  9th  "  30  " 
October,  lOth  "  31  " 
November,  11th  "  30  " 
December,  12th        "       31     " 


REDUCTION  OF  COMPOUND  NUMBERS.  97 

1.  Eeduce  2  hr.  to  seconds.  7200  sec. 

2.  Eeduce  7  da.  to  minutes.  10080  min. 

3.  Reduce  1  da.  3  hr.  44  min.  3  sec.  to  seconds. 

99843  sec. 

4.  Reduce  9  wk.  6  da.  10  hr.  40  min.  to  minutes. 

100000  min. 

5.  Reduce  1  mon.  3  da.  4  min.  to  minutes. 

44644  min. 

6.  Reduce  10800  seconds  to  hours.  3  hr. 

7.  Reduce  432000  seconds  to  days.  5  da. 

8.  Reduce  7322  seconds  to  hours.      2  hr.  2  min.  2  sec. 

9.  Reduce  4323  minutes  to  days.  3  da.  3  min. 

10.  Reduce  20280  minutes  to  weeks.  2  wk.  2  hr. 

11.  Reduce  41761  min.  to  months.      1  mo.  1  da.  1  min. 

MISCELLANEOUS     TABLES. 

I.  MEASURES  OF  WEIGHT. 

72.     Troy  Weight   is   used   in  weighing    gold,   silver, 
and  jewels. 

24  grains  (gr.)       make  1  pennyweight,  marked  pwt. 
20  pennyweights       "       1  ounce,  '^         oz. 

12  ounces  "       1  pound,  '^         lb. 

The  Standard  Unit  of  all  weight  in  the  United  States  is  the  Troy 
pomid,  containing  5760  grains. 

Apothecaries   "Weight    is    used    only    in    compounding 
medicines. 

20  grains  (gr.)  make  1  scruple,  marked  9. 

3  scruples  ''  1  dram,  "  3. 

8  drams  '^  1  ounce,  ''  5. 

12  ounces  "  1  pound,  "  lb. 


98  RAY'S  NEW  PKACTICAL  AKITHMETIC. 

The  following  are  also  used  by  apothecaries : 

60  minims  (or  drops)  rT\^.  make  1  fluid  drachm,  marked  f.  3. 

8  fluid  drachms  "      1  fluid  ounce,  "         f.  §. 

10  fluid  ounces  '*      1  pt.  (octarius)       "  O. 

8  pints  ''      1  gal.  (congius)      "       cong. 

II.  MEASURES  OF  LENGTH. 

The  following  measures  are  often  mentioned  and  most 
of  them  are  still  used  in  special  professions: 


12  lines  =1  inch. 

3  barleycorns  =  1  inch 

4  inches  =  1  hand. 
9  inches  = 1  span. 


3    feet  =  1  pace. 

6    feet  =1  fathom. 

3    miles  =  1  league. ' 

69^  miles  (nearly)  =  1  degree. 


Surveyors  use  a  chain  four  rods  long,  divided  into 
links  of  Ty^^  inches  each. 

Engineers  divide  the  foot  into  tenths  and  hundredths. 
The  yard  is  also  divided  similarly  in  estimating  duties 
at  the  custom  houses. 

A  degree  is  divided  into  60  nautical  or  geographic 
miles. 

A  nautical  mile  or  knot  is,  therefore,  nearly  1^  com- 
mon miles. 

Circular  Measure  is  used  in  measuring  circles. 

60  seconds  C)  make  1  minute,  marked  '. 
60  minutes  ^'      1  degree,  "         °. 

360  degrees  "      1  circle. 

Rem. — The  circumference  is  also  divided  into  quadrants  of  90° 
each,  and  into  siqjis  of  80°  each. 


KEDUCTION  OF  COMPOUND  NUMBERS.  99 

III.    MISCELLANEOUS  TABLE. 

12  things  make  1  dozen,  marked  doz. 

12  dozen         "  1  gross,  "         gr. 

12  gross  "  1  great  gross. 

20  things        "  1  score. 

100  pounds  of  nails,   make  1  keg. 
196  pounds  of  flour         "      1  barrel. 

200  pounds  of  pork  or  beef  make  1  barrel. 
240  pounds  of  lime  "      1  cask. 

24  sheets  of  paper  make  1  quire. 
20  quires  "      1  ream. 

2  reams  "      1  bundle. 

A  sheet  folded  in 

2  leaves  is  called  a     folio. 
4       '•        ''        "       a     quarto,  or  4to. 
8       "        ''        ''       an  octavo,  or  8vo. 
12       ''        "        "       a     duodecimo,  or  12mo. 
16       "        "        '^       a     16mo. 

Examples  in  Miscellaneous  Tables. 

73.     1.  Reduce  5  lb.  4  oz.  Troy  to  ounces.  64. 

2.  Eeduce  9  lb.  3  oz.  5  pwt.  to  pwt.  2225. 

3.  Eeduce  8  lb.  9  oz.  13  pwt.  17  gr.  to  gr.  50729. 

4.  Eeduce  805  pwt.  to  pounds.          3  lb.  4  oz.  5  pwt. 

5.  Eeduce  12530  gr.  to  pounds.     2  lb.  2  oz.  2  pwt.  2  gr. 

6.  Eeduce  4  lb.  5  g  2  gr.  to  grains.  25442. 

7.  Eeduce  7  lb.  2  §  1  9  to  grains.  41300. 

8.  Eeduce  431  3  to  pounds.                       4  lb.  5  5  7  5. 

9.  Eeduce  975  9  to  pounds.          •             3  lb.  4  §  5  5. 

10.  Eeduce  6321  gr.  to  pounds.       1  lb.  1  g  1  3  1  9  1  gr. 

11.  Eeduce  4  cong.  7  f  g  to  fluid  drams.  4152. 


100  KAY'S  NEW  PRACTICAL  AKITHMETIC. 

12.  Eediice  5  O.  6  f.  §  3  f.  3  to  minims.  41460. 

13.  Keduce  !H69  f.  5  to  gallons. 

2  cong.  3  O.  4  f.  3  5  f.  3. 

14.  Reduce  3  yd.  to  barleycorns.  324. 

15.  How  many  lines   in  1  foot  6  inches?  216. 

16.  What  is  the  height  of  a  horse  of  16^  hands? 

5  ft.  6  in. 

17.  A  field  measures  24  chains  in  length  and  15  chains 
in  breadth:   how  many  acres  in  it?  36. 

18.  A  cistern  contains  267  cubic  feet  624  cubic  inches : 
how  many  gallons  does  it  hold?     (Art.  64,  Eem.).      2000. 

19.  Reduce  8°  41'  45"  to  seconds.  31305. 

20.  Reduce  61°  59'  28"  to  seconds.  223168. 

21.  Reduce  915'  to  degrees.  15°  15'. 

22.  Reduce  3661"  to  degrees.  1°  1'  1". 

23.  What  cost  6  gross  of  screws  at  5  cents  a  dozen? 

$3.60. 

24.  A  man  is  4  score  and  10:  how  old  is  he?         90  yr. 

25.  At  18  certs  a  quire,  what  will  3  bundles  of  paper 
cost?  $21.60. 

26.  How  many  sheets  of  paper  will  be  required  for  a 
a  12mo.  book  of  336  pages?  14. 

27.  An  octavo  work  in  5  volumes  has  512  pages  in 
Vol.  1,  528  in  Yol.  2,  528  in  Vol.  3,  512  in  Vol.  4,  and 
496  in  Yol.  5:  how  much  paper  was  used  for  one  copy 
of  the  whole  work?  6  quires  17  sheets. 


Promiscuous    Examples. 

74.     1.  What  cost  2  bu.  of  plums,  at  5  ct.  a  pint? 

$6.40. 

2.  What   cost   3    bu.    2    pk.    of  peaches,    at    50    ct.    a 

l>eck?  $7. 


KEDUCTION  OF  COMPOUND  NUMBERS.  101 

3.  What  cost  3  pk.  3  qt.  of  baVl^y, 'ai  3*  ct.  a  pint? 

4.  At  15  ct.  a  peck,  how  many  bashels  of  apples  can 
be  bought  for  83?  5  bii. 

5.  If  salt  cost  2  ct.  a  pint,  how  much  can  l)e  bought 
with  SI. 66?  1  bu.  1  pk.  1  qt.  1  pt. 

6.  I  put  91  bu.  of  wheat  into  bags  containing  3  bu.  2 
pk.  each  :  how  many  bags  were  required  ?  26. 

Rem. — Reduce  both  quantities  to  pecks,  and  then  divide. 

7.  How  many  spikes,  weighing  4  oz.  each,  are  in  a 
parcel  weighing  15  lb.  12  oz.  ?  63. 

8.  I  bought  44  cwt.  52  lb.  of  cheese ;  each  cheese 
weighed  9  lb.  15  oz. :  how  many  cheeses  did  I  buy? 

448. 

9.  How  many  kegs,  of  84  lb.  each,  can  be  filled  from 
a  hogshead  of  sugar  weighing  14  cwt.  28  lb.?  17. 

10.  How  many  boxes,  containing  12  lb.  each,  can  be 
filled  from  7  cwt.  56  lb.  of  tobacco?  63. 

11.  If  a  family  use  3  lb.  13  oz.  of  sugar  a  week,  how 
long  will  6  cwt.  10  lb.  last  them?  160  wk. 

12.  What  will  2  acres  125  square  rods  of  land  cost,  at 
20  cents  a  square  rod?  $89. 

13.  A  farmer  has  a  field  of  16  A.  53  sq.  rd.  to  divide 
into  lots  of  1  A.  41  sq.  rd.  each :  how  many  lots  will  it 
make?  13. 

14.  How  many  cu.  in.  in  a  block  of  marble  2  ft.  long, 
2  ft.  high,  2  ft.  wide?  13824. 

15.  One  cu.  ft.  of  water  weighs  1000  oz.  avoirdupois: 
what  do  5  cu.  ft.  weigh?  312  lb.  8  oz. 

16.  What  is  the  weight  of  a  quantity  of  water  occupy- 
ing the  space  of  1  cord  of  wood,  each  cubic  foot  of  water 
weighing  1000  ounces  avoirdupois?  4  T. 


102  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

17'  jV  cufei6'  fo(>t I  af! '  oak  weighs  950  oz.  avoirdupois: 
Ay.bat  >l5c>"3  ;Coi\lw,*of;  oak'\\:eigh?  7  T.  1:^  cwt. 

*'  18*  \^ihd' the *eo»t^  of' 63' gallons  of  wine,  at  20  cents  a 
pint.  $100.80. 

19.  Find  the  cost  of  5  barrels  of  molasses,  each  con- 
taining 31  gal.  2  qt.,  at  10  cents  a  quart.  $63. 

20.  At  5  cents  a  pint,  what  quantity  of  molasses  can 
be  bought  for  $2?  5  gal. 

21.  How  many  dozen  bottles,  each  bottle  holding  3 
qt.  1  pt.,  can  be  filled  from  63  gal.  of  cider?  6  doz. 

22.  How  many  kegs,  of  4  gal.  3  qt.  1  pt.  each,  can  be 
filled  from  58  gal.  2  qt.?  12. 

23.  If  a  human  heart  beat  70  times  a  minute,  how 
many  times  will  it  beat  in  a  da}^?  100800. 

24.  How  many  seconds  in  the  month  of  February, 
1876?  2505600  sec. 

25.  If  a  ship  sail  8  miles  an  hour,  how  many  miles 
will  it  sail  in  3  wk.  2  da.  3  hr.  ?  4440  mi. 

26.  A  horse  is  fed  1  peck  of  oats  daily.  If  oats  cost 
44  cents  a  bushel,  how  much  will  it  cost  to  feed  him  a 
year  of  365  days?  $40.15. 

27.  A  flour  dealer  bought  40  barrels  of  flour  for  3 
ct.  a  pound,  and  sold  it  for  5  ct.  a  pound:  how  much 
did  he  gain?  $156.80. 


ADDITION  OP  COMPOUND  NUMBERS. 

75.  When  the  numbers  to  be  added  are  compound, 
the  operation  is  called  Addition  of  Compound  Numbers. 

1.  A  farmer  sold  three  lots  of  wheat:  the  first  lot 
contained  25  bu.  3  pk. ;  the  second,  14  bu.  2  pk. ;  the 
third,  32  bu.  1   pk. :  how  much  did  he  sell? 


ADDITION  OP  COMPOUND  NUMBERS. 


103 


Solution. — Place  units  of  the  same  denomination  in 
the  same  column  (Art.  17).  Beginning  with  pecks, 
and  adding,  the  sum  is  6,  which  is  reduced  to  bushels 
by  dividing  by  4,  the  number  of  pecks  in  a  bushel,  and 
there  being  2  pecks  left,  write  the  2  under  the  column 
of  pecks,  and  carry  the  1  bushel  to  the  column  of 
bushels;  adding  this  to  the  bushels,  the  sum  is  72,  which 
write  under  the  column  of  bushels. 


OPERATION. 

bu. 

pk. 

25 

3 

14 

2 

32 

1 

72 


(2) 

(3) 

bu. 

pk.       qt. 

pt. 

bu. 

pk. 

qt. 

pt, 

3 

2         0 

1 

7 

3 

7 

1 

4 

0         6 

1 

6 

2 

0 

0 

1 

3        7 

1 

9 

2 

4 

1 

"9 

2         6 

1 

24 

0 

4 

0 

Rule. — 1.  Write  the  numbers  to  be  added,  placing  units 
of  the  same  denomination  in  the  same  column. 

2.  Begin  with  the  lowest  denomination,  add  the  numbers, 
and  divide  their  stim  by  the  number  of  units  of  this  denom- 
ination ivhich  make  a  unit  of  the  next  higher. 

3.  Write  the  remainder  under  the  column  added,  and 
carry  the  quotient  to  the  next  column. 

3.  Proceed,  in  the  same  manner  with  all  the  colurnns  to 
the  last,  under  which  write  its  entire  sum. 

Proof. — The  same  as  in  Addition  of  Simple  Numbers. 


Rem.  1. — In  writing  compound  numbers,  if  any  intermediate  de- 
nomination is  wanting,  supply  its  place  with  a  cipher. 

Rem.  2. — In  adding  simple  numbers  we  carr}^  one  for  every  ten, 
because  ten  units  of  a  lower  order  always  make  one  of  the  next 
higher;  but,  in  compound  numbers,  the  scale  varies,  and  we  carry 
one  for  the  number  of  the  lower  order,  which  makes  one  of  the  next 
higher. 


104  RAY'S  NEW  PKACTICAL  AKITHMETIC. 


Examples 

DRY 

MEASURE. 

(4) 

(5) 

bu. 

pk. 

qt. 

bu. 

pk.   qt. 

pt. 

4 

3 

7 

8 

1    7 

1 

5 

2 

2 

7 

3    2 

1 

7 

1 

6 

9 

2    7 

1 

LIQUID    MEASURE. 


(6) 

(7) 

qt. 

pt. 

gi- 

gal. 

qt.   pt. 

gi- 

7 

1 

3 

40 

3    1 

3 

6 

0 

2 

16 

1    0 

2 

9 

1 

3 

71 

2    1 

2 

AVOIRDUPOIS  WEIGHT. 

(8)  (9) 

T.  cwt.   lb.   oz.           cwt.  lb.  oz. 

45   3   53   10           16  85  14 

14   14   75   15           15  90  13 

19   17   18   13           18  74  12 


LONG  MEASURE. 

(10) 

(11) 

mi.    rd. 

yd. 

ft. 

in. 

28    129 

4 

2 

11 

64    280 

3 

1 

9 

17    275 

5 

1 

8 

ADDITION  OF  COMPOUND  NUMBERS.  105 


SQUARE     MEASURE. 


(12) 
A.    sq.  rd. 
41      51 

sq.  yd. 
15 

(13) 
sq.  ft. 

8 

sq.  in. 
115 

64     104 

20 

7 

109 

193     155 

14 

5 

137 

CUBIC  MEASURE. 

(14)  (15) 

C.  cu.  ft.  cu.  in.       cu.  yd.  cu.  ft.  cu.  in. 

13   28    390          50  18    900 

15   90    874          45  17    828 

20   67    983          46  20    990 


TIME     MEASURE. 

(16)  (17) 

da.     Ill*,  min.  sec.  mo.  wk.  da.  lir.  min.  sec. 

16     18      28     47  3       0      0     23     51  40 

13     15      49     59  12      4     19     30  37 

19     16      53     42  3       1      5     13     27  18 

18.  Five  loads  of  wheat  measured  thus :  21  bu.  3  pk. ; 
14  bu.  1  pk. ;  23  bu.  2  pk. ;  18  bu.  1  pk. ;  22  bu.  1  pk. : 
how  many  bushels  in  all?  100  bu. 

19.  A  farmer  raised  of  oats  200  bu.  3  pk. ;  barley, 
143  bu.  1  pk. ;  corn,  400  bu.  3  pk. ;  wheat,  255  bu.  1 
pk. :  how  much  in  all?  1000  bu. 

20.  A  grocer  sold  5  hogsheads  of  sugar :  the  first 
weighed  8  cwt.  36  lb. ;  the  second,  4  cwt.  64  lb. ;  the 
third,  5  cwt.  19  lb. ;  the  fourth,  7  cwt.  75  lb. ;  the  fifth, 
7  cwt.  84  lb. :  what  did  all  weigh?  33  cwt.  78  lb. 

21.  Add  13  lb.  11  oz.;  17  lb.  13  oz. ;  14  lb.  14  oz. ;  16 
lb. ;   19  lb.  7  oz. ;   and  17  lb.  9  oz.  99  lb.  6  oz. 


106  RAY'8  NEW  PRACTICAL  ARITHMETIC. 

22.  Two  men  depart  from  the  same  place :  one  travels 
104  mi.  50  rd.  due  east;  the  other,  95  mi.  270  rd.  due 
west:  how  far  are  they  apart?  200  mi. 

23.  A  man  has  3  farms:  in  the  first  are  186  A.  134 
sq.  rd. ;  in  the  second,  286  A.  17  sq.  rd. ;  in  the  third, 
113  A.  89  sq.  rd. :  how  much  in  all?         586  A.  80  sq.  rd. 

24.  Add  17  sq.  yd.  3  sq.  ft.  119  sq.  in.;  18  sq.  yd.  141 
sq.  in. ;  23  sq.  yd.  7  sq.  ft. ;  29  sq.  yd.  5  sq.  ft.  116  sq. 
in.  88  sq.  yd.  8  sq.  ft.  88  sq.  in. 

25.  A  has  4  piles  of  wood  :  in  the  first,  7  C.  78  cu.  ft. ; 
the  second,  16  C.  24  cu.  ft.;  the  third,  35  C.  127  cu.  ft.; 
the  fourth,  29  C.  10  cu.  ft.:  how  much  in  all? 

88  C.  Ill  cu.  ft. 

26.  I  sold  4642  gal.  3  qt.  1  pt.  of  wine  to  A ;  945  gal. 
to  B;  1707  gal.  1  pt.  to  C;  10206  gal.  1  qt.  to  D:  how 
many  hogsheads  of  63  gal.  each  did  I  sell  ? 

277  hogsheads  50  gal.  1  qt. 


SUBTRACTION  OF  COMPOUND  NUMBERS. 

76.  When  two  given  numbers  are  compound,,  the 
operation  of  finding  their  difference  is  called  Subtraction 
of  Compound  Nximhers. 

1.  I  have  67  bu.  2  pk.  of  wheat:  how  much  will  re- 
main after  selling  34  bu.  3  pk.  ? 

Solution.  —  Write  the  less  number  under  the 
greater,  placing  units  of  the  same  denomination  in 
the  same  column.  3  pk.  can  not  be  taken  from  2  pk., 
but  1  bu.  being  taken  from  67  bu.  reduced  to  pk., 
and  added  to  the  2  pk.,  gives  6  pk.  3  pk.  from  6  pk. 
leaves  3  pk.;  34  bu.  from  66  bu.  leaves  32  bu.  The  "32  3~ 
difference  is,  therefore,  32  bu.  3  pk.    ^_ 


PERJl 

bu. 

lTION. 

pk. 

67 

2 

34 

3 

SUBTRACTION  OF  COMPOUND  NUMBERS.         107 

Rem. — Instead  of  diminishing  the  67  bu.  by  1,  the  result  will  be 
the  same  to  increase  the  lower  number  34  bu.  by  1,  as  is  done  in 
subtraction  of  simple  numbers. 

(2)  (3) 

bu.       pk.     qt.      pt.  bu.     pk.     qt.      pt. 

From  12       0       1       0  5       0       0       0 

Take    J 2^ 1  1       0 0_  1^ 

3       1       7       1  3       3       7       1 

Rule. — 1.  Write  the  less  number  under  the  greater,  placing 
units  of  the  same  denomination  in  the  same  column. 

2.  Begin  with  the  lowest  denomination,  and,  if  possible. 
take  the  lower  number  from  the  one  above  it. 

3.  But,  if  the  lower  number  of  any  denomination  be  greater 
than  the  upper,  increase  the  upper  number  by  as  many  units 
of  that  denomination  as  make  one  of  the  next  higher ;  sub- 
tract as  before,  and  carry  one  to  the  lower  number  of  the 
next  higher  denomination. 

4.  Proceed   in   the   same  manner  with  each  denomination. 

Proof. — The  same  as  in  Subtraction  of  Simj^le  Num- 
bers. 

Rem. — The  resemblance  between  subtraction  of  simple,  and  ^^f 
compound  numbers,  is  the  same  as  in  Addition  75,  Rem.  2. 


Examples, 
liquid  measure. 

gal.    qt.     pt.  gal.       qt.       pt.      gi. 

From  17       2       1  43         1         1         2 

Take    Hi       3       0,  23         3         1         3 


108 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


AVOIRDUPOIS    WEIGHT. 


(6) 
T.  cwt.  lb. 
From  14  12  50 
Take  10  13  75 


T.  cwt.  lb.  oz. 

16  7  18  14 

5  6  75  15 


LONG   MEASURE. 


(8) 
mi.        rd. 
From  18         198 
Take    11         236 


(9) 

yd.      ft.  in. 

4         1  10 

2         1  11 


SQUARE   MEASURE. 


( 

10) 

(11) 

A. 

8q.  rd. 

sq.  yd. 

sq.  ft. 

sq.  in. 

From  327 

148 

19 

6 

72 

Take  77 

155 

CUBIC 

16 

MEASURE. 

6 

112 

( 

12) 

(13) 

C. 

cu.  ft. 

cu.  yd. 

cu.  ft. 

cu.  in. 

From  28 

116 

18 

7 

927 

Take  19 

119 

TIME 

(14) 

9 

MEASURE. 

15 
(15) 

928 

hr. 

min.  sec. 

da. 

hr.  min, 

.  sec. 

From  18 

43  27 

245 

17   40 

37 

Take  17 

51  45 

190 

11   44 

42 

SUBTKACTION  OF  COMPOUND  NUMBEKS.         109 

16.  If  2  bu.  1  pk.  1  qt.  be  taken  from  a  bag  containing 
4  bushels  of  hickory  nuts,  what  quantity  will  remain  ? 

1  bu.  2  pk.  7  qt. 

17.  From  100  bu.  take  24  bu.  1  pt. 

75  bu.  3  pk.  7  qt.  1  pt. 

18.  I  bought  46  lb.  4  oz.  of  rice :  after  selling  19  lb.  8 
oz.,  how  much  remained?  26  lb.  12  oz. 

19.  A  wagon  loaded  with  hay  weighs  32  cwt.  66  lb. ; 
the  wagon  alone  weighs  8  cwt.  67  lb. :  what  is  the  weight 
of  the  hay?  23  cwt.  99  lb. 

20.  It  is  24899  miles  round  the  earth :  after  a  man 
has  traveled  100  mi.  41  rd.  what   distance  will   remain? 

24798  mi.  279  rd. 

21.  I  had  a  farm  containing  146  A.  80  sq.  rd.  of  land. 
I  gave  my  son  86  A.  94  sq.  rd. :  how  much  was  left  ? 

59  A.  146  sq.  rd. 

22.  From  8  C.  50  cu.  ft.  of  wood,  3  C.  75  cu.  ft.  are 
taken:  how  much  is  left?  4  C.  103  cu.  ft. 

23.  A  cask  of  wine  containing  63  gal.  le^J^ed;  only  51 
gal.  1  qt.  2  gi.  remained :  how  much  was  lost  ? 

11  gal.  2  qt.  1  pt.  2  gi. 

24.  From  5  da.  10  hr.  27  min.  15  sec.  take  2  da.  4  hr. 
13  min.  29  sec.  3  da.  6  hr.  13  min.  46  sec. 

77.  In  finding  the  time  between  any  two  dates,  con- 
sider 30  days  1  month,  and  12  months  1  year. 

1.  A  note,  dated  April  14,  1875,  was  paid  February 
12,  1877 :  find  the  time  between  these  dates. 

Solution. — In   writing   the    dates,    observe  operation. 

that   February  is   the   2d  month  of  the  year  yr.      mon.  da. 

and  April  the  4th;  then,  from  1877  yr.  2  mo.  187  7      2      12 

12  da.  subtract  1875  yr.  4  mo.  14  da.     The  re-  18  7  5      4      14 

mainder  is  1  vr.  9  mo.  28  da.  19      2  8 


110  KAY'S  NEW  PKACTICAL  ARITHMETIC. 

2.  The  Independence  of  the  United  States  was  declared 
July  4,  1776 :  what  length  of  time  had  elapsed  on  the 
Ist  of  September,  1876?  100  yr.  1  mo.  27  da. 

3.  The  first  crusade  ended  July  15,  1099;  the  third 
crusade,  July  12,  1191 :  find  the  difference  of  time  be- 
tween these  dates.  91  yr.  11  mo.  27  da. 

4.  Magna  Charta  was  signed  June  15,  1215 ;  Mary, 
Queen  of  Scots,  was  beheaded  February  8,  1587 :  find  the 
difference  of  time  between  these  dates. 

371  yr.  7  mo.  23  da. 

5.  The  battle  of  Hastings  was  fought  Oct.  14,  1066; 
William,  Prince  of  Orange,  landed  at  Tor  Bay  Nov.  5, 
1688:  what  w^as  the  difference  of  time  between  the 
two  events?  622  yr.  21   da. 

6.  The  battle  of  Austerlitz  was  fought  December  2, 
1805;  the  battle  of  Waterloo,  June  18,  1815:  find  the 
difference  of  time.  9  yr.  6  mo.  16  da. 

78.     To  fiijd  the  time  between  two  dates  in  days. 

1.  Find  the  number  of  days  from  May  10  to  Oct.  21. 


Solution.  — Of  May,  there  remains  31  —  10  == 
21  days;  there  are  30  days  in  June,  31  in  July, 
31  in  August,  30  in  September,  and  21  in  Octo- 
ber; then  the  number  of  days  from  May  10  to 
October  21,  is  21  -f  30  +  31  +  31  +  30  +  21  == 
164. 


2.  Find  the  number  of  days  from  March  17  to  Septem- 
ber 12.  179. 


OPERATION. 

31 

May, 

10 
21 

June, 

30 

July, 
Aug., 
Sept., 
Oct., 

31 
31 
30 
21 

164 

MULTIPLICATION  OF  COMPOUND  NUMBEKS.      HI 

3.  A  note  dated  April  18,  1877,  is  due  June  20,  1877: 
how  many  days  does  it  run?  63. 

4.  A   note    dated    Sept.    5,  1877,  is    due   Dec.  7,  1877 : 
how  many  days  does  it  run?  93. 

5.  Find   the    number   of  days    from    Oct.    12,  1877,  to 
May  25,  1878.  225. 

6.  Find    the    number   of  days  from    Aug.  20,  1875,  to 
March  8,  1876.  201. 


MULTIPLICATION  OF  COMPOUND  NUMBERS. 

79.  When  the  multiplicand  is  a  compound  number, 
the  operation  is  called  Multiplication  of  Compound  Numbers. 

1.  A  farmer  takes  to  mill  5  bags  of  wheat,  each  con- 
taining 2  bu.  3  pk. :  how  much  had  he  in  all  ? 

Solution. — Begin   at   the  lowest   denomination    for 

convenience.     Multiply  the  3  pk.  by  5,  making  15  pk.,  bu.     pk. 

which,  reduced,  gives  3  bu.  and  3  pk.;   write  the  3  pk.  2       3 

under  the  pecks,  and  carry  the  3  bu.     Then,  multiply  5 

the  2  bu.  by  5,  add  to  the  product  the  3  bu.,  and  write  13       3 
the  13  bu.  under  the  bushels. 

Rule. — 1.  Write  the  multiplier  under  the  lowest  denom- 
ination of  the  multiplicand. 

2.  Multiply  the  loivest  denomination  first,  and  divide  the 
product  by  the  number  of  units  of  this  denomination  ivhich 
make  a  unit  of  the  next  higher,  write  the  remainder  under 
the  denomination  multiplied,  and  carry  the  quotient  to  the 
product  of  the  next  higher  denomination. 

3.  Proceed  in  like  manner  with  all  the  denominations, 
writing  the  entire  product  at  the  last. 

Proof. — The  same  as  in  Simple  Multiplication. 


112  llA^'S  NEW  PRACTICAL  ARITHMETIC. 

Rem. — There  are  two  differences  between  multiplication  of  simple 
and  of  compound  numbers:  1.  In  simple  numbers  it  is  more  con- 
venient to  use  one  figure  of  the  multiplier  at  a  time;  in  compound 
numbers  it  is  better  to  use  the  eritire  tnultiplier  each  time.  2.  In 
simple  numbers  the  scale  is  miiforvi;  in  compound  numbers  it  varies 
with  the  table. 

Examples. 

2.  Multiply  2  bii.  1  pk.  1  qt.  1  pt.  by  6. 

13  bu.  3  pk.  1  qt. 

3.  Multiply  2  bu.  2  pk.  2  qt.  by  9.  23  bu.  2  qt. 

4.  If  4  bu.  3  pk.  3  qt.  1  pt.  of  wheat  make  1  bl.  of 
iiour,  how  much  will  make  12  bl.?         58  bu.  1  pk.  2  qt. 

5.  Find  the  weight  of  9  hogsheads  of  sugar,  each 
weighing  8  cwt.  62  lb.  3  T.  17  cwt.  58  lb. 

6.  How  much  hay  in  7  loads,  each  weighing  10  cwt. 
89  lb.?  3  T.  16  cwt.  23  lb. 

7.  If  a  ship  sail  208  mi.  176  rd.  a  day,  how  far  will 
it  sail  in  15  days?  3128  mi.  80  rd. 

8.  Multiply  23  cu.  yd.  9  cu.  ft.  228  cu.  in.  by  12. 

280  cu.  yd.  1  cu.  ft.  1008  cu.  in. 

9.  Multiply  16  cwt.  74  lb.  by  119.        99  T.  12  cwt.  6  lb. 

10.  Multiply  47  gal.  3  qt.  1  pt.  by  59. 

2824  gal.  2  qt.  1  pt. 

11.  A  travels  27  mi.  155  rd.  in  1  day:  how  far  will 
he  travel  in  one  month  of  31  days?  852  mi.  5  rd. 

12.  In  17  piles  of  wood,  each  pile  containing  7  C.  98 
cu.  ft. :  what  is  the  quantity  of  wood?       182  C.  2  cu.  ft. 

13.  Multiply  2  wk.  4  da.  13  hr.  48  min.  39  sec.  by 
75.  49  mo.  3  wk.  3  hr.  48  min.  45  sec. 

14.  A  planter  sold  75  hogsheads  of  sugar,  each  weigh- 
ing 10  cwt.  84  lb.,  to  a  refiner,  for  6  ct.  a  pound.  The 
refiner  sold  the  sugar  for  8  ct.  a  pound :  how  much  did 
he  gain?  $1626. 


DIVISION  OF  COMPOUND  NUMBEES.  113 

15.  A  cotton-factor  sold  425  bales  of  cotton,  each 
weighing  4  cwt.  85  lb.,  for  13  ct.  a  pound.  He  paid 
$24735  for  the  cotton  :  how  much  did  he  gain  ? 

$2061.25. 

DIVISION  OF  COMPOUND  NUMBERS. 

80.  When  the  dividend  is  a  compound  number,  the 
operation  is  called  Division  of  Compound  Numbers. 

The  divisor  mvij  be  either  a  Simple  or  a  Compound 
Number.     This  gives  rise  to  two  cases: 

First. — To  find  how  often  one  Compound  Number  is 
contained  in  another  Compound  Number. 

This  is  done  by  reducing  both  divisor  and  dividend  to  the  .^ame 
denomination  before  dividing  (Examples  6  and  8,  Art.  74). 

Second. — To  divide  a  Compound  Number  into  a  given 
number  of  equal  parts.  This  is  properly  Compound  Di- 
vision. 

1.  Divide  14  bu.  2  pk.  1  qt.  by  3. 

Solution. — Divide   the  highest   denomination  operation. 

first,  so  that,  if  there  be  a  remainder,  it  may  be  re-         bu.     pk.     qt. 
duced  to  the  next  lower  denomination,  and  added     8)14       2        1 
to  it.     3  in  14  is  contained  4  times,  and  2  bu.  are  4       3       3 

left;  write  the  4  under  the  bushels,  and  reduce 
the  remaining  2  bu.  to  pk.,  to  which  add  the  2  pk.,  making  10  pk. 
This,  divided  by  3,  gives  a  quotient  of  3  pk.,  with  1  pk.  remaining; 
which,  reduced  to  qt.,  and  1  qt.  added,  gives  9  qt.     This,  divided  by 
3,  gives  a  quotient  3,  wdiich  is  written  under  the  quarts. 

(2)  (3) 

bu.    pk.    qt.  da.      hr.    min.    sec. 

7)33     2     6  5)17     12     56     15 

4     3     2  3     12     11     15 

PPAC.  8. 


114  KAY'S  NEW  riiACTICAL  ARITHMETIC. 

Rule. — 1.  Write  the  quantity  to  be  divided  in  the  order 
of  its  denominations,  beginning  with  the  highest;  place  the 
divisor  on  the  left, 

2.  Begin  ivith  the  highest  denomination,  divide  each  num- 
ber  separately,  and  write  the  quotient  beneath. 

3.  If  a  remainder  occurs  after  any  division,  reduce  it  to 
the  next  lower  denomination,  and,  before  dividing,  add  to  it 
the  number  of  its  denomination. 

Proof. — The  same  as  in  Simple  Division. 

Rem.— Each  partial  quotient  is  of  the  same  denomination  as 
that  part  of  the  dividend  from  which  it  is  derived. 

4.  Divide  67  bu.  3  pk.  4  qt.   1  pt.  by  5. 

13  bu.  2  pk.  2  qt.   1  pt. 

5.  Eleven  casks  of  sugar  weigh  35  ewt.  44  lb.  12  oz. : 
what  is  the  average  weight  of  each? 

3  cwt.  22  lb.  4  oz. 

t).  I  traveled  39  mi.  288  rd.  in  7  hr. :  at  what  rate  per 

hour  did  I  travel?  5  mi.  224  rd. 

7.  Divide  69  A.  64  sq.  rd.  by  16.  4  A.  54  sq.  rd. 

8.  490  bu.  2  pk.  4  qt. -^  100.  4  bu.  3  pk.  5  qt. 

9.  265  lb.  10  oz. -^50.  5  lb.  5  oz. 

10.  45  T.  18  cwt.-^17.  2  T.  14  cwt. 

11.  114  da.  22  hr.  45  min.  18  sec. -1-54. 

2  d«a.  3  hr.  5  min.  17  sec. 

12.  10  cwt.  27  lb.  13  oz. -f-23.  44  lb.  11  oz. 

13.  309  bu.  2  pk.  2  qt.-^78.  3  bu.  3  pk.  7  qt. 

14.  127  gal.  3  qt.  1  pt.  3  gi.  — 63.  2  gal.  1  gi. 

15.  788  mi.  169  rd.  -^319."  2  mi.  151  rd. 
16    A  farmer  has  two  farms,  one  of  104  A.  117  sq.  rd.; 

the  other,  87  A.  78  sq.  rd.  He  reserves  40  A.  40  sq.  rd., 
and  divides  the  remainder  equally  among  his  3  sons : 
what  is  the  share  of  each  son?  50  A,  105  sq.  rd. 


LONGITUDE  AND  TIME.  115 

17.  A  farmer's  crop  consisted  of  5000  bu.  3  pk.  of  corn 
one  year,  and  7245  bu.  2  pk.  the  year  following.  He  sold 
B022  bu.  1  pk.  and  placed  the  remainder  in  8  cribs, 
each  crib  containing  an  equal  amount :  how  many  bushels 
in  each  crib?  528  bu. 

18.  A  speculator  bought  6  adjoining  pieces  of  land, 
each  containing  4  A.  80  sq.  rd.  He  divided  the  whole 
into  54  lots,  and  sold  them  at  $5  a  sq.  rd. :  how  much 
did  he  get  for  each  lot?  S400. 

19.  Add  35  lb.  9  oz..  75  lb.  14  oz.,  85  lb.  15  oz.;  from 
the  sum  take  186  lb.  14  oz. ;  multiply  the  remainder  by 
8;  divide  the  product  by  64:  what  is  the  result? 

1  lb.  5  oz. 

LONGITUDE    AND    TIME. 

81.  Difference  of  longitude  and  time  between  different 
1^  laces. 

The  circumference  of  the  earth,  like  other  circles,  is 
divided  into  360  equal  parts,  called  degrees  of  longitude. 

The  sun  appears  to  pass  entirely  round  the  earth, 
360°,  once  in  24  hours,  one  day ;  and  in  1  hour  it  passes 
over  15°.     (360° -f- 24  .:=  15°). 

As  15°  equal  900',  and  1  hour  equals  60  minutes  of 
time,  therefore,  the  sun  in  1  minute  of  time  passes  over 
l^'  of  Si  degree,     (900'~60=r=:  15'). 

As  15'  equal  900",  and  1  minute  of  time  equals  60 
seconds  of  time,  therefore,  in  1  second  of  time  the  sun 
passes  over  15"  of  a  degree.     (900"^  60  =  15"). 

Table  for  Comparing  Longitttde  and  Time. 

15°  of  longitude -=.- 1  hour  of  time. 
15'  of  longitude  ^^^  1  min.  of  time. 
15"  of  longitude  =  1  sec.     of  time. 


116  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

1.  How  many  hr.  min.  and  sec.  of  time  correspond  to 
18°  25'  :W  of  longitude?  1  hr.  13  min.  42  sec. 

Analysis. — By  inspection  of  the  table,  it  is  evident  that, 

Degrees  (  °  )  of  longitude,  divided  by  15,  give  hours  of  time. 
Minutes  ( ^  )  of  longitude,  divided  by  15,  give  minutes  of  time. 
Seconds  (^^)  of  longitude,  divided  by  15,  give  seconds  of  time. 

Hence,  if  18°  25^  30^^  of  Ion.  be  divided  by  15,  the  quotient  will 
be  the  time  in  hr.  min.  and  sec.  corresponding  to  that  longitude. 

To  find  the  time  corresponding  to  any  difference  of 
longitude : 

Rule. — Divide  the  longitude  by  15,  according  to  the  rule 
for  Division  of  Compound  Numbers ,  and  mark  the  quotient 
hr.  min.  sec,  instead  of    °     '     ". 

Conversely:  To  find  the  longitude  corresponding  to 
any  difference  of  time. 

Rule. — Multiply  the  time  by  15,  according  to  the  rule 
for  Multiplication  of  Compound  Numbers^  and  mark  the 
product     ®     '    "    instead  of  hr.  rain.  sec. 

2.  The  difference  of  longitude  between  two  places  is  30*^ : 
what  is  their  difference  of  time?  2  hr. 

8.  The  difference  of  longitude  betw^een  two  places  is  71° 
4':  what  is  the  difference  of  time?        4  hr.  44  min.  16  sec. 

4.  The  difference  of  longitude  between  New  York  and 
Cincinnati  is  10°  35':  what  is  the  difference  of  time? 

42  min.  20  sec. 

5.  The  difPerence  of  time  between  Cincinnati  and  Phil- 
adelphia is  37  min.  20  sec. :  what  is  the  difference  of 
longitude?  9°  20'. 


LONGITUDE  AlNl)  TIME.  117 

6.  The  difference  of  time  between  !N^ew  York  and  St. 
Louis  is  1  hr.  4  min.  56  sec. :  what  is  the  difterence  of 
longitude?  16°  14^ 

7.  The  difference  of  time  between  London  and  Wash- 
ington is  5  hr.  8  min.  4  sec. :  what  is  the  difference  of 
longitude?  77°  I'o 

DlB^FERENCE  IN   TiME. 

82.  It  is  noon  (12  o'clock),  at  any  i)l^ce  when  the 
sun  is  on  the  meridian  of  that  place. 

As  the  sun  appears  to  travel  from  the  east  toward  the 
west,  when  it  is  noon  at  any  place,  it  is  after  noon  east 
of  that  place,  and  before  noon  icest  of  that  place. 

Hence,  a  place  has  later  or  earlier  tirne  than  another, 
according  as  it  is  east  or  west  of  it.     Therefore, 

When  the  time  at  one  place  is  given,  the  time  at  another, 
if  EAST  of  this,  is  found  by  adding  their  difference  of  time; 
if  WEST,  by  SUBTRACTING  their  difference  of  time. 

8.  When  it  is  noon  at  Cincinnati,  what  is  the  time  at 
Philadelphia?  37  min.  20  sec.  past  noon. 

9.  When  it  is  11  o'clock  A.  M.  at  New  York,  what  is 
the  time  in  longitude  30°  east  of  New  York?         1  P.  M. 

10.  When  12  o'clock  (noon)  at  Philadelphia,  what  is 
the  time  at  Cincinnati?  11  hr.  22  min.  40  sec.  A.  M. 

11.  When  it  is  11  o'clock  A.  M.  at  New  York,  what  is 
the  time  at  St.  Louis?  9  hr.  55  min.  4  sec.  A.  M. 

12.  Wheeling,  W.  Ya.,  is  in  longitude  80°  42'  west: 
the  mouth  of  the  Columbia  river, in  longitude  124°  west: 
when  it  is  1  o'clock  P.  M.  at  Wheeling,  what  is  the  time 
at  the  mouth  of  Columbia  river? 

10  hr.  6  min.  48  sec.  A.  M. 


DEFINITIONS. 

83.  1.  Factors  of  a  number  are  two  or  more  num- 
bers, the  product  of  which  equals  the  given  number 
(Art.  28,  2). 

Thus,  2  and  3  are  factors  of  6,  because  2X3  =  6;  2,  3,  and  6  are 
factors  of  30,  because  2  X  3  X  ^  =  30- 

Kem.  1. — One  and  the  number  itself  are  not  considered  factors  of  a 
number. 

Kem.  2. — A  number  may  be  the  product  of  more  than  one  set  of 
factors.     Thus,  2  X  6==  12,  3  X  4  =  12,  and  2X2X^  =  12. 

2.  A  multiple  of  a  number  is  a  product  of  which  the 
number  is  a  factor. 

Thus,  6  is  a  multiple  of  3;  30  is  a  multiple  of  5. 

3.  INTumbers  are  divided  into  two  classes,  prime  and 
composiie. 

4.  A  prime  number  has  no  factors. 

Thus,  5,  11,  17  are  prime  numbers. 

5.  A  composite  number  has  two  or  more  factors. 

Thus,  6,  12,  30  are  composite  numbers. 

6.  A  prime  factor  is  a  factor  which  is  a  prime 
p  umber. 

Thus,  3  is  a  prime  factor  of  12. 
(118) 


FACTORING.  119 

7.  A  factor  is  common  to  two  or  more  numbers  when 
it  is  a  factor  of  each  of  them. 

Thus,  3  is  a  common  factor  of  12  and  15. 

Rem. — Sometimes  the  smallest  of  two  or  more  numbers  may  be 
the  common  factor.     Thus,  6  is  a  common  factor  of  6,  12,  and  18. 

8.  Two  or  more  numbers  are  prime  to  each  other ^  w^hen 
they  have  no  common  factor. 

Thus,  9  and  10  are  prime  to  each  other. 

9.  A  common  divisor  (C.  D.)  of  two    or   more  num- 
bers is  any  common  factor. 

Thus,  2,  3,  and  6  are  each  a  common  divisor  of  12  and  18. 

10.  The   greatest   common   divisor  (G.  C.  D.)  of  two 
or  more  numbers  is  the  greatest  common  factor. 

Thus,  6  is  the  greatest  common  divisor  of  12  and  18. 

11.  A  common  multiple  (C.  M.)  of  tw^o  or  more  num- 
bers is  any  multiple  of  all  of  them. 

Thus,  6,  12,  18,  etc.,  are  common  multiples  of  2  and  3. 

12.  The   least  common  multiple  (L.  C.  M.)  of  two  or 
more  numbers  is  the  least  multiple  of  all  of  them. 

Thus,  6  is  the  least  common  multiple  of  2  and  3. 

13.  Factoring    is    the    process   of  resolving    composite 
numbers  into  their  factors. 


To  Find  the  Prime  Jfiniihers. 

84.     All    the   prime    numbers    except    2  are  odd  num- 
bers. 


120  RAYS  NEW  PKACTICAL  ARITHMETIC. 

Rule. — 1.  Write  the  odd  tuanUrs  in  a  series  1,  ;5,  5,  7, 
1),  etc. 

2.  After  3  eixtse  every  'M  number;  after  5  erase  every 
bth  number ;  after  7  erase  every  7th  number;  after  11  erase 
every  llth  number y  etc. 

3.  Then  2  and  the  numbers  that  remain  are  the  prime 
numbers. 

KxERCiSE. — Find  the  prime  numbers  from  1   to  100. 

85.  The  operations  of  Factoring  depend  upon  the 
following 

PRINTIPI.ES. 

1.  A  factor  of  a  number  exactly  divides  it. 
Tlius,  5  is  a  factor  of  30  and  is  contained  in  it  G  times. 

2.  A  multiple  of  a  number  exactly  contains  it. 
Thus,  30  is  a  multiple  of  5  and  contains  it  G  times. 

3.  A  factor  of  a  number  is  a  factor  of  any  midtiple  of 
that  number. 

Thus,  3  being  a  factor  of  6  is  a  factor  of  12,  18,  24,  etc. 

4.  A  composite  number  is  equal  to  the  product  of  all  its 
prime  factors. 

Thus,  the  prime  factors  of  30  are  2,  3,  and  5;  2  X  ^  X  ^  =  30. 

86.  In  resolving  numbers  into  their  prime  factors  it 
will  be  found  convenient  to  remember  the  following 
facts  in  reference  to  the  prime  numbers  2,  3,  and  5. 


FACTORING.  121 

1.  Two  is  a  factor  of  every  even  number. 
Thus,  2  is  a  factor  of  4,  6,  8,  10,  etc. 

2.  Three  is  a  factor  of  a  number  when  the  sum  of  its 
digits  is  3  or  some  multiple  of  3. 

Thus,  3  is  a  factor  of  2457;  for  2  -[-  4  +  5  +  7  =  18,  which  is  6 
times  3. 

3.  Five  is  a  factor  of  every  number  ichose  unit  figure  is 
0  or  5. 

Thus,  5  is  a  factor  of  10,  15,  20,  25,  etc. 

Rem. — Whether  the  prime  numbers  7,  11,  13,  etc.,  are  factors  of  a 
number  or  not  is  best  ascertained  by  trial. 

To  Resolve  a  J^umher  into  its  Prime  Factors. 
87.     1.  Resolve  30  into  its  prime  factors. 

Solution.— 2  is  a  factor  of  30  (Art.  86,  1).     Di- 
viding 30  by  2,  the  quotient  is  15.     3  being  a  factor  of  operation. 
15  (Art.  86,  2)  is  also  a  factor  of  30  (Art.  85,  Prin.  2)30 
3).     Dividing  15  by  3  the  quotient  is  5,  a  prime  num-  3)15 
ber.     Then,  2,  3  and  5  are  the  prime  factors  of  30.  5 

Rule. — 1.  Divide  the  given  number  by  any  prime  number 
that  unll  exactly  divide  it. 

2.  Divide  the  quotient  in  the  same  manner ;  and  so  con- 
tinue to  divide,  until  a  quotient  is  obtained  which  is  a 
prime  number. 

3.  The  several  divisors  and  the  last  quotient  will  be  the 
prime  factors  of  the  given  number. 

Rem. — It  will  be  most  convenient  to  divide  each  time  by  the 
smallest  prime  number. 


122 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Resolve  the  following  into  their  prime  factors: 


2. 

4. 

^    2 

^1   -J. 

23. 

39. 

3,  13 

3. 

8. 

2,  2,  2. 

24. 

40. 

2,  2,  2,  5 

4. 

9. 

3,  3. 

25. 

42. 

2,  3,  7 

5. 

10. 

2,  5. 

26. 

44. 

2,2,11 

6. 

12. 

2,  2,  3. 

27. 

45. 

3,  3,  5 

7. 

14. 

2,  7. 

28. 

46. 

2,  23 

8. 

15. 

3,  5. 

29. 

48. 

2,  2,  2,  2,  3 

9. 

16. 

9 

■^5 

2,  2,  2. 

30. 

49. 

7,  7 

10. 

18. 

2,  3,  3. 

31. 

50. 

2,  5,  5 

11. 

20. 

2,  2,  5. 

32. 

70. 

2,  5,  7 

12. 

22. 

2,  11. 

33. 

77. 

7,  11 

13. 

24. 

9 

2,  2,  3. 

34. 

91. 

7,  13 

14. 

25. 

5,  5. 

35. 

105. 

3,  5,  7 

15. 

26. 

2,  13. 

36. 

119. 

7,  17 

16. 

27. 

3,  3,  3. 

37. 

133. 

7,  19 

17. 

28. 

2,  2,  7. 

38. 

154. 

2,  7,  11 

18. 

32. 

2  2 

2^2 

39. 

210. 

2,  3,  5,  7 

19. 

34. 

2/17; 

40. 

231. 

3,  7,  11 

20. 

35. 

5,  7. 

41. 

330. 

2,  3,  5,  11 

21. 

36. 

2 

2,  3,  3. 

42. 

462. 

2,  3,  7,  11 

22. 

38. 

2,  19. 

43. 

2310. 

2,  3,  5,  7,  11 

88.     To    find    the    prime    factors    common    to    two   or 
more  numbers. 


1.  What  prime  factors  are  common  to  30  and  42? 

Solution. — Write  the  numbers  in  n  line.     2  is  a 

prime  factor  of  both  30  and  42  (Art.  86,  1).     Di-  operation. 

viding  by  2,  the  quotients  are  15  and  21.     3  is  a  2)30    4  2 

prime  factor  of  both  15  and  21  (Art.  86,  2);   and  3)15   Tl 

consequently  of  both  30  and  42  (Art.  86,  Prin.  3).  5        7 
Dividing  by  3,  the  quotients  5  and  7  are  prime  to 
each  other  (Art.  83,  8).     Then  2  and  3  are  the  common  factors. 


FACTORING.  123 

Rule.—  1.    Wiite  the  given  mimbers  in  a  line. 

2.  Divide  by  any  prime  number  that  will  exactly  divide  all 
oj  them;  divide  the  quotients  in  the  same  manner ;  and  so 
continue  to  divide  until  two  or  more  of  the  quotients  are 
prime  to  each  other. 

3.  Then  the  several  divisors  will  be  the  common  factors. 

What  prime  factors  arc  common  to 

2,  3,  5. 

9      9     9 

— ,    -J,    -J. 

2,  2,  3. 

2,  3,  3. 

3,  3,  3. 
2,  2,  5, 
2,  3,  7. 

2,  2. 

2,  3. 

3,  3. 

2,  5. 

3,  5. 
5,  5. 
2,  7. 

11. 
13. 
17. 
19. 
23. 

89.  Finding  the  G.  C.  D.  of  two  or  more  numbers 
depends  upon  the  following 

Principle. —  The  G.  C.  D.  of  two  or  more  numbers  con- 
tains all  the  prime  factors  common  to  the  numbers,  and  no 
other  factor. 


2. 

60 

and  90? 

3. 

56 

and  88? 

4. 

72 

and  84? 

5. 

54  and  90? 

6. 

81 

and  108? 

7. 

80 

and  100? 

8. 

84  and  126? 

9. 

52, 

68  and  76? 

10. 

66, 

78  and  102? 

11. 

63, 

99  and  117? 

12. 

50, 

70  and  110? 

13. 

45, 

75  and  105? 

14. 

75, 

125  and  175? 

15. 

42, 

70  and  98? 

16. 

33, 

55,  77  and  121? 

17. 

39, 

65,  91  and  104? 

18. 

34, 

51,  85  and  102? 

19. 

38, 

57,  95  and  114? 

20. 

46, 

69,  92  and  115? 

124  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

Thus,  the  G.  C.  D.  of  12  and  18  is  6;  it  contains  the  common  fac- 
tors 2  and  3;  it  must  contain  both  of  them,  else  it  would  not  be  the 
greatest  C.  D.;  it  can  contain  no  other  factor,  else  it  would  not  divide 
both  12  and  18. 

1.  Find  the  G.  C.  D.  of  80  and  42. 

First  Method. 

OPERATION. 

Solution. — The  prime  factors  common  to  30  and  2)30  42 
42  are  2  and  3  (Art.  88);  their  product  is  6;  then  3)15  21 
the  G.  C.  D.  of  30  and  42  is  6  (  Prin.).  5        7 

Rule. — 1.  F'uid  the  prime  factors  common  to  the  given 
numbers. 

2.  Multiply  them  together. 

3.  The  product  will  be  the  greatest  common  divisor. 

Second  Method. 

Solution. — Dividing  42  by  30,  the  re-^ 
mainder  is  12;   dividing  30  by  12,  the  re-  operation. 

mainder  is  6;    dividing   12  by  6,  the  re-       30)42(1 
mainder  is  0.     Then  6  is  the  G.  C.  D.  of  3  0 

30  and  42.     For,  30  r=  6  X  5  a"d  42  =  6  T^)  3  0  (  2 

X  7;   then,  because  5  and  7  are  prime  to  2  4 

each  other,  6  must  contain  all   the  prime  0)12(2 

factors  common  to  30  and  42;  it- is,  there-  12 

fore,  their  G.  C.  D.  (Prin.). 

Rule. — 1.  Divide  the  greater  number  by  the  less,  the 
divisor  by  the  remainder,  and  so  on,  always  dividing  the 
last  divisor  by  the  last  remainder,  until  nothing  remains. 

2.   The  last  divisor  nill  be  the  greatest  common  divisor. 

Rem. — To  find  the  G.  C.  D.  of  more  than  two  numbers,  first  find 
the  G.  C.  D.  of  two  of  them,  then  of  that  common  divisor  and  one 
of  the  remaining  numbers,  and  so  on  for  all  the  numbers;  the  last 
common  divisor  will  be  the  G.  C.  D.  of  all  the  numbers. 


FACTORING.  125 

Find    the   greatest    common    divisor    of  the    following 
numbers : 


12. 

18. 
20. 
27. 
30. 
16. 
24. 
36. 
31. 
26. 
23. 
19. 
17. 
39. 
.227. 
12. 

5. 

8. 


90.  Finding  the  L.  C.  M.  of  two  or  more  innnbers 
depends  upon  the  following     * 

Principle.— TAe  L.  C.  M.  of  two  or  more  numbers  con- 
tains  all  the  prime  factor's  of  each  number  and  no  other 
factor. 

Thus,  the  L.  C.  M.  of  12  and  18  is  J^6;  its  prime  factors  arc  %  %  3, 
and  8;  it  must  contain  all  these  factors,  else  it  would  not  contain 
both  the  numbers;  it  must  contain  no  other  factor,  else  it  would  not 
he  the  least  CM. 


2. 

16,  24  and  40. 

3. 

24,  36  and  60. 

4. 

36,  54  and  90. 

5. 

40,  60  and  100. 

6. 

54,  81  and  108. 

7. 

60,  90  and  120. 

8. 

32,  48,  80  and  112. 

9. 

48,  72,  96  and  120. 

10. 

72,  108,  144  and  180. 

11. 

62  and  93. 

12. 

78  and  130. 

13. 

161  and  253. 

14. 

247  and  323. 

15. 

391  and  697. 

16. 

2145  and  3471. 

17. 

16571  and  38363. 

18. 

72,  120  and  132. 

19. 

75,  125  and  165. 

20. 

64,  96,  112  and  136. 

126 


EAY'S  NEW  PRACTICAL  ARITHMETIC. 


1.  Find  the  L.  C.  M.  of  4,  6,  9  and  12. 


Solution. — The  prime  factors  of  4  are 
2  and  2;  those  of  6  are  2  and  3;  of  9,  3 
and  3;  and  of  12,  2,  2,  and  3.  Then,  the 
prime  factors  of  the  L.  C.  M.  are  2,  2,  3, 
o,  and  no  other  factor  ( Prin.).  Hence, 
36  is  the  L.  C,  M. 


OPERATION. 

4  =  2X2 
6  =  2X3 
9  =  3X3 
12  =  2X2X3 

2X2X3X3  =  36 


The  process  of  factoring  and  selecting 
the  prime  factors  for  the  L.  C.  M.  is  very 
much  simplified  hy  the  operation  in  the 
form  of  Short  Division,  as  shown. 


OPERATION. 

2)4    6    9     12 
2 


12     3     9 


6 


3)3     9        3 


2  X  2  X  3  X  3  =  36. 


Rule. — 1.    Wii'te  the  given  numbers  m  a  line, 

2.  Divide  by  any  prime  munher  that  will  exactly  divide 
two  or  more  of  them. 

3.  Write  the  quotients  and  undivided  numbers  in  a  line 
beneath, 

4.  Divide  these  numbers  in  the  same  manner^  and  so  con- 
tinue the  operation  until  a  line  is  reached  in  which  the 
numbers  are  all  prime  to  each  other. 

5.  Then  the  product  of  the  divisors  and  the  numbers  in 
the  last  line  will  be  the  least , common  multiple. 

Rkm. — "When  the  quotient  is  1  it  need  not  be  written. 


Find  the  least  common  multiple  of 


2. 

4, 

6  and  8. 

3. 

6, 

9  and  12. 

4. 

4, 

8  and  10. 

5. 

6, 

10  and  15. 

6. 

6, 

8,  9  and  12. 

24. 
36. 
40. 
30. 

72. 


CANCELLATION.  127 

7.  10,  12,  15  and  20.  60. 

8.  9,  15,  18  and  30.  90. 

9.  12,  18,  27  and  36.  108. 

10.  15,  25,  30  and  50.  150. 

11.  14,  21,  30  and  35.  210. 

12.  15,  20,  21  and  28.  420. 

13.  20,  24,  28  and  30.  840. 

14.  45,  30,  35  and  42.  630. 

15.  36,  40,  45  and  50.  1800. 

16.  42,  56  and  63.  504. 

17.  78,  104  and  117.  936. 

18.  125,  150  and  200.  3000. 

19.  10,  24,  25,  32  and  45.  7200. 

20.  2,  3,  4,  5,  6,  7,  8  and  9.  2520. 

21.  16,  27,  42,  and  108.  3024. 

22.  13,  29,  52,  and  87.  4524. 

23.  120,  360,  144,  720,  and  72.  720. 


CANCELLATION. 

91.  1.  I  bought  3  oranges  at  5  cents  each,  and  paid 
for  them  with  pears  at  3  cents  each  :  how  many  pears 
did  it  take? 

OPERATION, 

Solution  I.— 5  cents   multiplied  by  3  are  15  5 

cents,  the  price  of  the  oranges.     15  divided  by  3  3 

is  5,  the  number  of  pears.  3)15 

5 

From  a  consideration  of  this  example  and  its  solution 
we  have  the  following 

Principle.— A  mimber  is  not  changed  by  multiplying  it 
and  then  dividing  the  product  by  the  mtdtipUer. 


128 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


For    the    example,  then,    we    may  offer    the    following 
solution  and  operation  : 


Solution  II. — Indicate  the  multiplica- 
tion and  division;  then,  erase  or  cancel 
the  multiplier  3  and  the  divisor  3  hy 
drawing  a  line  across  them;  and  write 
the  result,  equal  to  5. 


OPERATION. 


Rem. — The  product  5  X  ^  forms  a  dividend  of  which  3  is  the 
divisor. 

2.  If  1  buy  10  pears  at  3  cents  each,  and  pay  for  them 
with  oranges  at  5  cents  each :  how  many  oranges  will  it 
take  ? 


Solution. — 5  is  a  factor  of  10,  for  10 
=  5X2;  then,  cancel  the  divisor  5  and 
also  the  factor  5  'in  10  by  canceling  10 
and  writing  the  remaining  factor  2 
above  it.  The  product  of  the  remaining 
factors  is  6. 


OPERATION. 


2 


—==6 


3.  Divide  15  X  21  by  14  x  10. 


Solution. — 5  is  a  common  factor  of  15 
and  10;  then,  cancel  15,  writing  3  above 
it,  and  10,  writing  2  below  it.  7  is  a 
common  factor  of  14  and  21;  then,  cancel 
14,  writing  2  below  it,  and  21,  writing  3 
above  it.  The  product  of  the  factors  re- 
maining in  the  dividend  is  9,  and  of  those 
remaining  in  the  divisor  is  4;  the  quotient 
of  9  divided  by  4  is  2\.     Therefore, 


OPERATION. 

3        3 
2         2 


--f=-2i 


Cancellation  is  a  process  of  abbreviation  by  omitting 
the  common  factors  of  the  dividend  and  divisor. 


CANCELLATION.  129 

Rule. — 1 .  Cancel  the  factors  common  to  both  the  dividend 
and  divisor. 

2.  Divide  the  product  of  the  factors  remaining  in  the 
dividend  by  the  product  of  the  factors  reynainijig  in  the 
divisor. 

3.  The  result  will  be  the  quotient  required. 

4.  How  many  barrels  of  molasses,  at  S13  a  barrel,  will 
pay  for  13  barrels  of  flour,  at  $4  a  barrel  ?  4. 

5.  Multiply  17  by  18,  and  divide  by  6.  51. 

6.  In  15  times     8,  how  many  times  4?  30. 

7.  In  24  times     4,  how  many  times  8?  12. 

8.  In  37  times  15,  how  many  times  5?  111. 

9.  Multiply  36  by  40,  and  divide  the  product  by  30 
multiplied  by  8.  6. 

10.  In  36  times  5,  how  man}^  times  15?  12. 

11.  Multiply  42,  25,  and  18  together,  and  divide  the 
product  by  21  X  15.  60. 
^  12.  I  sold  23  sheep,  at  $10  each,  and  was  paid  in  hogs, 

at  $5  each  :  how  many  did  I  receive  ?  46. 

y  13.  How  many  yards  of  flannel,  at  35  cents  a  yard, 
will  pay  for  15  yards  of  calico,  at  14  cents?  6  yd. 

V  14.  What  is  the  quotient  of  21  X  H  X  6  X  26,  divided 
by  13X3X14X2?  33. 

^  15.  The  factors  of  a  dividend  are  21,  15,  33,  8,  14, 
and  17 ;  the  divisors,  20,  34,  22  and  27 :  required  the 
quotient.  49. 

V  16.  I  bought  21  kegs  of  nails  of  95  pounds  each,  at  6 
cents  a  pound ;  paid  for  them  with  pieces  of  muslin  of 
35  yards  each,  at  9  cents  a  yard :  how  many  pieces  of 
muslin  did  I  give?  38. 

17.  What  is  the  quotient  of  35  X  39  X  -10  divided  by 
26  X  30  X  42  ?  If. 

Prac.  9. 


130   *        KAY'S  NEW  PKACTICAL  ARITHMETIC. 

18.  What   is   the  quotient  of  26  X  ^3  X  35  divided  by 
4X9X25?  -  33^'. 

19.  What   is   the  quotient  of  6  X  9  X  15  ><  21  divided 
by  4X6X10X14?  S^^. 

20.  What  is  the  quotient  of  21  X  24  X  28  X  35  divided 
by  14  X  18X20X25?  3||. 


<33?^»Jt'/^]f^ 


FRACTIONS. 


'    "%!'<* 


92.  A  unit  may  be  divided  into  equal  parts ;  thus, 

1st.  An  apple  may  be  divided  equally  between  two  boys,  by  cut- 
ting it  into  two  equal  parts. 

2d.  An  apple  may  be  divided  equally  among  three  boys,  by  cut- 
ting it  into  three  equal  jpartfi. 

3d.  In  like  manner,  an  apple  may  be  divided  into/o?/r,^ve,  six,  or 
any  number  of  equal  parts. 

These  equal  parts  into  which  a  unit  may  be  divided  are  called 
fractions. 

DEFINITIONS. 

93.  1.  A  fraction  is  one  or  more  equal  parts  of  a 
unit. 

2.  To  express  fractions  by  words  and  figures. 

When  a  unit  is  divided  into  two  equal  parts, 

Each  part  is  called  one-half,  written  i. 

Both  parts  are  called  iifo -halves,  "  f. 
When  a  unit  is  divided  into  three  equal  parts. 

Each  part  is  called  one-third,  written  ^. 

Two  parts  are  called  two-thirds,  "  f. 

All  the  parts  are  called  three-thirds,  "  f. 
When  a  unit  is  divided  into  four  equal  parts. 

Each  part  is  called  one-fourth,         written  \. 

Two  parts  are  called  two-fourths,         "  f. 

Three  parts  are  called  three-fourths,      "  f . 

All  the  parts  are  called  four -fourths,    "  |. 

(131) 


132  KAY'S  NEW  PKACTICAL  AK1THMP:T1C. 

When  a  unit  is  divided  into  five  equal  parts, 
Each  part  is  called  one-fifth,  written  \, 

Two  parts  are  called  two-fifths,  "         J. 

Three  parts  are  called  threefifths,        "         |. 
Four  parts  are  called /owr-//f As,  "         ^. 

All  the  parts  are  called  five-fifths,        "         |. 
When   a   unit   is   divided   into   six,    seven,   eight,  etc., 
equal  parts,  each  part   is   called   one-sixth,  ^,  one-seventh, 
|,  one-eighth,  \,  etc. 

94.  1.  A  fraction  is  expressed  in  words  by  two  num- 
bers; the  first  numbers  the  parts,  the  second  names  them; 
the  first  number  is  called  the  numerator,  the  second  is 
called  the  denominator. 

2.  A  fraction  is  expressed  in  figures,  by  writing  the 
numerator  above  the  denominator  with  a  line  between 
them. 

3.  The  numerator  and  denominator  are  styled  the 
terms  of  the  fraction. 

4.  The  denominator  shows  into  how  many  equal  parts 
the  unit  is  divided,  and  the  numerator,  how  many  of 
the  parts  are  taken. 

95.  When  a  unit  is  divided  into  equal  parts,  the  size 
of  each  part  depends  upon  the  number  of  the  parts. 

Thus,  if  apples  of  equal  size  be  divided,  one  into  two  equal  parts, 
another  into  three  equal  parts,  a  third  mio  four  equal  parts,  etc.,  a 
half  will  be  larger  than  a  third,  a  third  larger  than  a  fourth,  etc. 
Hence, 

1st.  The  less  the  number  of  parts  into  which  a  unit 
is  divided,  the  greater  the  size  of  each  part. 

2d.  The  greater  the  number  of  parts  into  which  a 
unit  is  divided,  the  less  the  size  of  each  part. 


COMMON  FKACTIONS.  I33 

96.     1.  A  fraction  may  also  be  regarded  as  a  part  of 
one  or  more  units. 


Thus: 

1st.  Two  apples  may  be  divided  equally  among  three  boys. 

Each  boy  will  receive,  either  one-third  of  each  of  the  two  apples, 
or  two-thirds  of  one  of  the  apples;  therefore,  ^  of  2  is  §.  Hence,  § 
may  be  considered  either  as  two-thirds  or  as  one-third  of  two. 

2d.  Two  apples  may  be  divided  equally  between  two  boys. 

Each  boy  will  receive,  either  one-half  of  each  of  the  two  apples,  or 
one  of  the  two  apples;  therefore,  ^  of  2  is  |,  or  1.  Hence,  |  may  be 
considered  either  as  two  halves  or  as  one-half  of  two. 

3d.  Three  apples  may  be  divided  equally  between  two  boys. 

Each  boy  will  receive,  either  one-half  of  each  of  the  three  apples, 
or  one  apple  and  one-half  of  another;  therefore,  J  of  3  is  |,  or  1^. 
Hence,  |  may  be  considered  either  as  three  halves  or  as  one-half  of 
three. 

2.  A  fraction  is  a  part  of  one  or  more  units. 

3.  The  numerator  expresses  the  ntimber  of  units. 

4.  The  denominator  expresses  the  part  of  each  to  be 
taken. 

97.  1.  A  fraction  may  also  be  regarded  as  an  ex- 
pression of  division,  in  which  the  numerator  is  the  dw- 
idend  and  the  denominator  the  divisor. 

Thus: 

1st.  f  is  2  divided  by  3;  here,  the  division  can  only  be  indicated, 

2d.  I  is  4  divided  by  2;  in  this  case,  the  division  can  be  performed 
exactly,  giving  a  quotient  2. 

3d.  f  is  5  divided  by  2;  in  this  case,  the  division  can  not  be  per- 
formed exactly,  the  quotient  being  2^. 

2.  A  fraction  is  an  indicated  division.  The  numerator 
is  the  dividend  and  the  denominator  is  the  divisor. 


134  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

3.  A  whole  number  may  be  expressed  in  tli('  form  of 
a  fraction,  by  writing  the  number  for  the  umiierator 
and  1  for  the  denominator. 

Thus,  2  may  be  written  |;  for  2  divided  by  1  is  2;  3  may  be  writ- 
ten |;  4  may  be  written  f,  etc. 

98.  The  value  of  a  fraction  is  its  relation  to  a  unit. 

1.  When  the  numerator  is  less  than  the  denominatoi-, 
the  value  of  the  fraction  is  less  than  1. 

Thus,  ^,  J,  §,  etc.,  arc  less  tlian  1. 

2.  When  the  numerator  is  equal  to  the  denominator, 
the  value  of  the  fraction  is  equal  to  1. 

Thus,  f,  |,  f,  etc.,  equal  1. 

3.  When  the  numerator  is  greater  than  the  denomina- 
tor, the  value  of  the  fraction   is  givater  than   1. 

Thus,  |,  |,  f,  etc.,  are  greater  than  1. 

4.  A  proper  fraction  is  one  whose  value  is  less  th'an  1. 

5.  An  improper  fraction  is  one  whose  value  is  equal 
to  or  greater  than  1. 

6.  A  mixed  number  is  a  whole  number  and  a  fraction. 

99.  1.  A  fraction  may  be  divided  into  equal  ])ai'ts. 

Thys,  after  an  apple  has  been  divided  into  two  equal  parts,  each 
half  may  be  divided  into  two  equal  parts;  the  whole  apple  will  then 
be  divided  into  four  equal  parts;  therefore,  ^  of  ^  is  \. 

Such  expressions  as   ^   of    ^,   ^  of   ^,   etc.,   are  termed  compoimd 

fractions. 

2.  A  compound  fraction  is  a  fraction  of  a  fraction. 


COMMON  FRACTIONS.  135 

100.  1.    Fractions    sometimes    occur    in     whicli    the 
numerator,  the  denominator  or  botli  are  fractional. 

Thus,  -r>  oT>  ^»  are  such  expressions;  they  are  called  complex 
fractions.     They  are  read  3^  divided  hy  4,  etc. 

2.  A  simple  fraction  is  one  in  which  both  terms  are 
entire. 

3.  A  complex  fraction  is  one  in  whi^h  one  or  both  of 
the  terms  are  fractional. 

101.  The  operations  with  fractions  depend  upon  the 
following 

'  Principles. 

1.  A  fraction  is  multiplied  by  multiplying  the  numerator. 

Thus,  if  the  numerator  of  |  be  multiplied  bj'  3,  the  result  will  be 
f ;  in  ^  the  parts  are  of  the  same  size  as  in  |,  but  there  are  three 
times  as  many. 


2.  A  fraction  is  divm^d  hy  dividing  the  numerator. 

Thus,  if  the  numerator  of  |  be  divided  hy  3.  the  result  will  be  ^; 
in  ^  the  parts  are  of  the  same  size  as  in  f ,  but  there  are  only  one- 
third  as  many. 

I 

3.  A  fraction  is  divided  hy  multiplying  the  denominator. 

Thus,  if  the  denominator  of  |  be  multiplied  by  3,  the  result  will 
be  |;  in  I  there  are  the  same  number  of  parts  as  in  |,  but  the  parts 
are  only  one-third  as  large. 

4.  A  fraction  is  multiplied  hy  dividing  the  denominator. 

Thus,  if  the  denominator  of  |  be  divided  by  3,  the  result  will  be 
|;  in  I  there  are  the  same  number  of  parts  as  in  |,  but  the  parts  are 
three  times  as  larsre. 


136  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

5.  Multiplying  both  terms  of  a  fraction  by  the  same  num- 
ber does  not  change  its  value. 

Thus,  if  both  terms  of  |  be  multiplied  by  2,  the  result  is  y«^;  in  ^ 
there  are  twice  as  many  parts  as  in  J,  but  they  are  only  one-half  as 
large. 

G.  Dividing  both  terms  of  a  fraction  by  the  same  number 
does  not  change  its  value. 

Thus,  if  both  terms  of  /^  be  divided  by  2,  the  result  will  be  |;  in 
I  there  are  only  one-half  as  many  parts  as  in  3^5,  but  they  are  twice  as 
large. 

These  six  j)rinciple8  may  be  stated  more  briefly,  as 
follows :  ^ 

I.  A  fraction  is  multiplied, 

1st.  By  multiplying  the  numerator. 
2(1.  By  dividing  the  denominator. 

II.  A  fraction  is  divided, 

1st.  By  dividing  the  numerator. 
2d.  By  7nultiplying  the  denominator. 

III.  The  value  of  a  fraction  is  not  changed, 

1st.   By  multiplying  both  terms  by  the  same  number. 
2d.  By  dividing  both  terms  by  the  same  number. 

The  operations  with  fractions  are  Beduction,  Addition, 
Subtraction,  Multiplication  and  Division. 


COMMON  FRACTIONS.  137 


REDUCTION  OF  FRACTIONS. 

102.     Reduction  of  Fractions  is  changing  their  form 
witiiout  altering;  their  value.     There  are  six  cases. 


CASE    I. 

103.     To  reduce   an   integer  to  an  improper  fraction, 
having  a  given  denominator. 

1.  In  3  apples,  how  many  halves? 

OPERATION. 

Solution. — In  1  apple  there  are  2  halves;  then,  in  |  X  3  =  f 

3  apples  there  are  3X2  halves  =  6  halves. 

Rule.—l.   Multiply  the  integer  by  the  given  denominator ; 
under  the  product  write  the  denominator. 

2.  In  4  apples,  how  many  halves? 

3.  In  2  apples,  how  many  thirds? 

4.  In  3  apples,  how  many  fourths? 

5.  In  4  apples,  how  many  fifths?  ^-f- 

6.  In  6  inches,  how  many  tenths  ?  fj 

7.  In  8  feet,  how  many  twelfths?  ff 

8.  Eeduce     4  to  sevenths.  t^ 

9.  Eeduce     8  to  ninths.  V' 

10.  Eeduce  19  to  thirteenths.  W 

11.  Eeduce  25  to  twentieths.  ^ 

12.  Eeduce  37  to  twenty-thirds.  W 


CASE    II. 

104r.     To    reduce    a    mixed    number   to    an    improper 

fraction. 


138  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

1.  In  31  apples,  how  many  halves? 

OPERATION. 

Solution. — In  1  apple  there  are  2  halves;  then,  in        |X3=| 
3  apples  there  are  3  X  ^  halves  =  6  halves.     6  halves 
and  1  half  are  7  halves.  1  +  1  =  1 

Rule. — 1.  Multiply  the  integer  by  the  denominator  of  the 
fraction;  to  the  product  add  the  numerator,  and  under  the 
sum  write  the  denominator. 

2.  In  4^  apples,  how  man}^  halves?  f 

3.  In  2\  apples,  how  many  thirds?  J 

4.  In  2|  apples,  how  many  thirds?  | 

5.  In  5 1  dollars,  how  many  fourths?  ^^- 

6.  Reduce       8J    to  an  improper  fraction.  ^- 

7.  Reduce     12f     to  an  improper  fraction.  -^ 

8.  Reduce     15f    to  an  improper  fraction.  ^ 

9.  Reduce  26^     to  an  improper  fraction.  ^^- 
It).  Reduce     3^     to  an  improper  fraction.  ^~^- 

11.  Reduce  46|       to  an  improper  fraction.  ^-p 

12.  Reduce  21^i|  to  an  improper  fraction.  ^"Ml^ 

13.  Reduce  lyVA  ^^  ^'^  improper  fraction.  |^^| 

14.  Reduce  14^^     to  an  improper  fraction.  "Ti^ 

15.  Reduce  lOy^  to  an  improper  fraction.  tVt 


CASE    ITT. 

105.     To  reduce  an  improper  fraction  to  an  integer  or 
mixed  number. 

1.  In  I  of  an  apple,  how  many  apples? 

OPERATION. 

Solution.— There  are  2  halves  in  1  apple;  then,  in  2)6 

6  halves,  there  are  6^2=3  apples.  ~3 


COMMON  FRACTIONS.  139 
2.  In  f  of  a  dollar,  how  many  dollars? 

OPERATION. 

Solution. — There  are  4  fourths  in  1  dollar;  then,  4  )  9 

in  9  fourths,  there  are  9  -=-  4  =;  2^  dollars.  2^ 

Rule. — 1.  Divide  the   numerator  by  the  denominator ;  the 
quotient  will  be  the  integer  or  the  mixed  number. 

3.  In    f    of  an  apple,  how  many  apples?  2. 

4.  In  -1^  of  an  apple,  how  many  apples?  3. 

5.  In  ^-^  of  a  dollar,  how  many  dollars?  $3|. 

6.  In   y^  of- a  dollar,  how  many  dollars?  S3f. 

7.  In  ^  of  a  bushel,  how  many  bushels?  2^  bu. 

8.  In  ^  of  a  dollar,  how  many  dollars?  ^^y^ij^. 

9.  In  -2^  of  an  ounce,  how  many  ounces?  8^  oz. 

10.  In  -5^  of  a  dollar,  how  many  dollars?  $131 

11.  Eeduce     ^^     to  a  mixed  number.  18|. 

12.  Reduce    ^\^    to  a  mixed  number.  15|. 

13.  Reduce    ^^    to  a  mixed  number.  25|^i. 

14.  Reduce  ^^^  to  an  integer.  40. 

15.  Reduce    ^^U^    to  an  integer.  31. 

16.  Reduce    ^j^-    to  a  mixed  number.  l^y^T* 

17.  Reduce    ^^    to  a  mixed  number.  46-j^. 

18.  Reduce  ^-^-^  to  a  mixed  number.  2im. 

19.  Reduce  ^-f^  to  a  mixed  number.  6^T%- 

20.  Reduce  ^^^-  to  an  integer.  199. 

21.  Reduce  -^^  to  a  mixed   number.  I^tot* 

CASE    IV. 

106.     To   reduce  a   fraction   to   higher  terms. 

A  fraction  is  reduced  to  higher  terms   by  multiplying 

both  terms  by  the  same  number.     This  does  not  change 
its  value  (Art.  101,  Prin.  5). 


140  RAYS  NEW  PRACTICAL  ARITHMETIC. 

1.  Eeduee  |  to  thirtieths. 

OPERATION. 

Solution. — 30  divided  by  5  is  6.     Multiplying        3  0-t-5=:    6 
both  terms  of  |  by  6,  the  result  is  ff  6X4  =  24 

Kule. — 1.  Divide  the  required  denominator  by  the  denom- 
inator of  the  given  fraction. 

2.  Multiply  both  terms   of  the  fraction   by  the  quotient; 
the  result  will  be  the  required  fraction. 


2. 

Eeduce 

i 

to 

fourths. 

f 

3. 

Reduce 

1 

to 

sixths. 

*• 

4. 

Reduce 

1 

to 

twelfths. 

T%- 

5. 

Reduce 

5 

to 

twenty -fourths. 

If- 

6. 

Reduce 

f 

to 

twenty-eighths. 

If 

7. 

Reduce 

A 

to 

eighty-fourths. 

if- 

8. 

Reduce 

i 

to 

seventy -seconds. 

n- 

9. 

Reduce 

f 

to  sixtieths. 

n- 

10. 

Reduce 

A 

to 

hundredths. 

- 1%- 

11. 

Reduce 

9 
2'U 

to 

a    fniction 

whose 

denominator    is 

720. 

T5ir- 

12. 

Reduce 

H 

to 

a    fraction 

whose 

denominator    is 

2016. 

im- 

13. 

Reduce 

H 

to 

a    fraction 

whose 

denominator    is 

1935. 

99  Q 
1935- 

14. 

Reduce 

If 

to 

a    fraction 

whose 

denominator    is 

8118. 

nn- 

15. 

Reduce 

If 

to 

a    fraction 

whose 

denominator    is 

5134. 

4832 
5134* 

16. 

Reduce 

H 

to 

a    fraction 

whose 

denominator    is 

2332^ 

I 

am- 

17. 

Reduce 

« 

to 

a    fraction 

whose 

denominator    is 

2541. 

mi- 

COMMON  FRACTIONS.  ^  141 

CASE    V. 

107.     To  reduce  a  fraction  to  its  lowest  terms. 

1.  A  fraction  is  reduced  to  lower  terms  by  dividing 
both  terms  by  the  same  number.  This  does  not  change 
its  value.     (Art.  101,  Prin.  6). 

2.  A  fraction  is  in  its  lowest  terms  when  the  numer- 
ator and  denominator  are  prime  to  each  other.  (Art. 
83,   8). 

1.  Eeduce  |^  to  its  lowest  terms. 

First    Method. 

Solution. — 2  is  a  common  factor  of  24  and  30  operation. 

(86,  1 ).     Dividing  both  terms  of  |^  by  2,  tbe  result  2  4  _  1  2 

is    -^f.      3   is  a  common   factor  of   12  and    15  (86,  ^~30~T5' 

2).     Dividing   both   terms   of  ^|  by  3,  the  result  is  o\  12         4 

f .     4  and  5  are  prime  to  each  other.  1  5  ~~    5 

Rule. — 1.  Divide'  both  terms  of  the  given  fraction  by  any 
common  factor. 

2.  Divide  the  resulting  fraction  in  the  same  manner. 

3.  So  continue  to  divide  until  a  fraction  is  obtained  whose 
terms  are  prime  to  each  other. 


Second  Method. 

OPERATION. 

24)30(1 
Solution. — The  greatest  common  divisor  of  2  4 

24  and  30  is  6.     Dividing  both  terms  of  |J  by  ~6~)  2  4(4 

6,  the  result  is  -|-.  2  4 

'30         5 


142 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Rule. — 1.  Divide  both  terms  of  the  given  fraction  by  their 
greatest  common  divisor. 

2.   The  resulting  fraction  will  be  in  its  lowest  terms. 


2. 

Eeduce 

u 

to  its  lowest  terms. 

i 

3. 

Eeduce 

n 

to  its  lowest  terms. 

I 

4. 

Reduce 

H 

to  its  lowest  terms. 

|. 

5. 

Reduce 

n 

to  its  lowest  terms. 

I 

6. 

Reduce 

T% 

to  its  lowest  terms. 

I 

7. 

Reduce 

n 

to  its  lowest  terms. 

I 

8. 

Reduce 

A-V 

to  its  lowest  terms. 

f- 

9. 

Reduce 

1% 

to  its  lowest  terms. 

if. 

10. 

Reduce 

m 

to  its  lowest  terms. 

i^T- 

11. 

Reduce 

m 

to  its  lowest  terms. 

if- 

12. 

Reduce 

m 

to  its  lowest  terms. 

If 

13. 

Reduce 

iV^ 

to  its  lowest  terms. 

A. 

14. 

Reduce 

i¥A 

to  its  lowest  terms. 

1^^. 

15. 

Reduce 

^m 

to  its  lowest  terms. 

.^T 

16. 

Reduce 

m 

to  its  lowest  terms. 

u- 

17. 

Reduce 

585 
1287 

to  its  lowest  terms. 

fV- 

18. 

Reduce 

Tiffs 

to  its  lowest  terms. 

A- 

19. 

Reduce 

mi 

to  its  lowest  terms. 

i¥r. 

20. 

Reduce 

mi 

to  its  lowest  terms. 

CASE    VI. 

m- 

108.     To  reduce  two   or  more  fractions  to  their  least 
common   denominator. 


1.  Two  or  more  fractions  have  a  common  denominator 
when  they  have  the  same  denominator. 

2.  A  common  denominator  of  two  or  more  fractions 
is   a    commx)n    multiple  of  their  denominators    (83,    11). 


COMMON  FRACTIONS. 


143 


3.  The  least  common  denominator  of  two  or  more 
fractions  is  the  least  common  multiple  of  their  denom- 
inators  (83,  12). 


1.  Eeduce 
nominator.    , 


I,  and 


to  their  least  common   de- 


OPERATION. 

2)4    6    9     12 


Solution. — The  least  common 
multiple  of  the  denominators  4, 
6,  9,  and  12  is  36  (90).  Each 
fraction,  then,  must  be  reduced 
to  thirty-sixths  (106).     i  =  U* 

5  3  0       8  3  2       nnc\      H   3  3 

-6  —  3^'    ¥  —  ^S'     ^^^^    T2  —  3T- 


2)2     3     9 


3)3     9 


2X2X3 

X  3  =  36. 

36^4r=    9 

36  -  6  =  .6 

9X3  =  27 

6  X  5  =  30 

4-  27 

?  3^ 

i=U 

86 -=-9=    4 

36 --12=    3 

4^X  8  =  32 

3X11  =  33 

l-=M 

ih  =  U 

Rule. — 1.  Find  the  L.  C.  M.  of  the  denominators  of  the 
fractions  for  their  least  common  denominator. 

2.  Reduce  each  fraction  to  another  having  this  denomin- 
ator. 

Kem.  1. — Integers  must  be  reduced  to  the  common  denominator 
by  Art.  103,  Rule. 

Rem.  2. — Before  commencing  the  operation,  mixed  numbers  must 
be  reduced  to  improper  fractions  (104). 

Rem.  3. — Each  fraction  must  be  in  its  lowest  terms  (107 ). 

Rem.  4.— Two  or  more  fractions  may  be  reduced  to  any  common 
denominator  in  the  same  way. 

Eeduce  to  their  least  common  denominator: 


2. 

3. 
4. 


h 

f- 

1- 

■l, 

*. 

1- 

i. 

f, 

*• 

A, 

T%, 

A- 

ii 

if, 

II- 

M- 

TO"? 

M- 

144  RAY'S  NEW  PKACTICAL  AKITHMETIC. 


5. 

.3 

I 

fV- 

6. 

i 

h 

J- 

7. 

3 

5 

5 

T' 

^' 

¥• 

8. 

h 

3 

i- 

9. 

I 

1. 

A- 

10. 

h 

f. 

H- 

11. 

h 

i 

i 

*• 

12. 

I 

I 

f. 

f- 

43. 

h 

A> 

A. 

a- 

14. 

I 

1. 

I 

H- 

15. 

2 

i, 

s 

A- 

16. 

2i 

f- 

4, 

5f. 

H. 

M, 

e- 

if, 

fl, 

M- 

M, 

M, 

n- 

1. 

t, 

h 

A, 

«, 

A- 

H, 

H, 

iJ- 

n, 

M- 

M, 

M- 

m, 

llf, 

fflf. 

m- 

A. 

if, 

if. 

ii- 

#i 

M, 

IS, 

M- 

H, 

If, 

il 

fj- 

faj 

i!, 

-¥/, 

■w- 

ft. 

M, 

li. 

n- 

17.     2i,    3i,    4i,    5. 

18-     A,    ii,     ii,     if.     If 

TTTfJ   TT¥>   Trf >   TTT'   ITT* 
!"•      T'         TTT'      TT>       85»      A>      Tff- 

TtVtT'    126"I5"'    12  6  0'   TTslj''    TJB^TF'   T^VlT- 
2^-      fj        iV'      A'      ii'      if'      ff- 

^T)T7'   71)^17'   ¥llir'   ^17'   ^TTTTj   luiT' 


ADDITION  OF  FRACTIONS. 

109.  Addition  of  Fractions  is  the  process  of  finding 
the  sum  of  two  or  more  fractional  numbers.  There  are 
two  cases. 

CASE    I. 

110.  When  the  fractions  have  a  common  denominator. 
1.  Add  i  I  and  f. 

OPERATION. 

Solution.— The  sum  of  1  fifth,  2  fifths,  and  3         i  _[-  2  _^  3  r=    6 
fifths,  is  6  fifths,     f  are  equal  to  1^  (Art.  106).  l  =  H 


COMMON    FRACTIONS.  145 

Explanation. — Since  the  denominators  are  the  same,  the  numer- 
ators express  parts  of  the  same  size;  therefore,  add  1  fifth,  2  fifths, 
and  3  fifths,  as  you  would  add  1  cent,  2  cents,  and  3  cents;  the  sum, 
in  one  case,  being  6  fifths,  in  Ihe  other,  6  cents. 

Rule.— 1.  Add  the  numerators;  under  the  sum  write  the 
common  denominator. 

Rem.  1 — The  result,  if  an  improper  fraction,  must  be  reduced  to  an 
integer,  or  a  mixed  number  (Art.  105). 

Rem.  2. — The  result  must  be  reduced  to  its  lowest  terms  (Art. 
107). 

3.  Add 

4.  Add     I,       I        3^       f  If 

5.  Add     4,       4,       i,       I.  22. 


h 

I> 

i- 

h 

h 

f, 

f 

h 

h 

h 

f- 

i, 

s 

h 

1- 

A, 

tV> 

A, 

H- 

5 
T3? 

A. 

A- 

H- 

tV. 

iV 

H, 

if- 

9 
20? 

TO? 

¥1 

H- 

fi      Add       -3_        _7_       _8_        iil.  9  6 


7.  Add    -i-V,    T%,    A-  H-  2t^ 

8.  Add    tV„    a.    H,  if-  2f 

QAHH-JL1113  17  91 

V.    ^UU        2  0?       "20"?        2U?  "2"0^-  ^"^^ 

10.  Add     if,     il     if,  ||.  2f 


CASE  II. 

111.     When  the  fractions  have  not  a  common  denom- 
inator. 

1.  Add  I,  I,  and  |i. 

Solution. — Reducing  the  fractions  operation. 

to  a  common  denominator  (Art.  108),  |  ^^ff     f  =  f f    ih  =  II 

I  =  f f^  f -=  fi  and  H  =  !l;  then,  the  ff  +  f |  +  H  =  || 

sum  of  f^,  ff,  and  |f  is  |f  .     ||  are  U^^U 
equal  to  2  f|. 

Explanation. — Since  the  denominators  are  different,  the  numer- 
ators do  not  express  parts  of  the  same  size;  therefore,  the  fractions 
can  not  be  added  till  they  are  reduced  to  a  common  denominator. 
Prac.  10. 


146 


RAYS  NEW  PRACTICAL  ARITHMETIC. 


Bule. — 1.  Reduce  the  fractions  to  a  common  denominator. 
2.  Add  the  numerators^  and  under  the  sum  write  the  com- 
mon denominator.  , 

Rem.  1. — Integers  and  fractions  may  bo  added  separately  and 
their  sums  then  united. 

Rem.  2. — The  integral  and  the  fractional  parts  of  mixed  numbers 
may  be  added  separately  and  their  sums  then  united. 


2. 

Add 

\  and 

i- 

3. 

Add 

\  and 

i- 

4.. 

Add 

\  and 

f. 

5. 

Add 

1  and 

*• 

6. 

Add 

1  and 

1- 

7. 

Add 

i  and 

+*. 

8. 

Add 

21   and 

31. 

A- 


Solution. — The  sum   of    \  and   |   is  |;    JrrrlJ;  operation. 

write  the  \  under  the  column  of  fractions  and  carry  2\ 

the  1  to  the  column  of  integers.     The  sum  of  1,  3,  3^ 

and  2  is  6.                                                                 •  6^  A7y.s. 


9.  Add 

10.  Add 

11.  Add 

12.  Add 

13.  Add 

14.  Add 

15.  Add 

16.  Add 

17.  Add 

18.  Add 

19.  Add 

20.  Add 

21.  Add 


^'  ¥' 


h 

h  W 

1 

1         2 

¥' 

J'   IT- 

i 

n.  8f- 

tV 

'  tV  Vt' 

iV- 

H 

.  A-  ii, 

M- 

tV 

,  2f,  3|, 

3f 

16 

1,  12f,  8f,  2i. 

h 

i  i.^i, 

i- 

I 

iV'   "JIT' 

i4«>  TirV^ 

tV 

'    16'    T2"' 

IfV,  2H- 

f. 

2i,  ^, 

6i,  8i. 

H 

,  4f  2i, 

2A- 

2\. 

2A\. 
10f|. 

1. 

21H- 
9f|. 


COMMON    I'KACTIONS.  U7 


SUBTRACTION  OF   FRACTIONS. 

112.  Subtraction  of  Fractions  is  the  process  of 
finding  the  difference  between  two  fractional  numbers. 
There  are  two  cases. 

CASE  I. 

113.  When  the  fractions  have  a  common  denomin- 
ator. 

1.  From  f  subtract  ^. 

OPERATION. 

Solution. — 2  'sevenths   from   5  sevenths   leaves   3       ^  —  7  =  7 
sevenths. 

Explanation. — Since  the  denominators  are  the  same,  the  num. 
erators  express  parts  of  the  same  size;  therefore,  subtract  2  sevenths 
from  5  sevenths  as  you  would  subtract  2  cents  from  5  cents;  the  re- 
mainder, in  one  case,  being  3  sevenths,  in  the  other  3  cents. 

Kule. — 1.  From  the  greater  numerator  subtract  the  less; 
under  the  remainder  write  the  common  denominator. 

2.  From     |  subtract    \.  \. 

3.  From     |  subtract     |.  \. 

4.  From     |  subtract     |.  \. 

5.  From  -^-^  subtract  f\.  \. 

6.  From  3^  subtract  If. 

operation. 
Solution. — f  can  not  be  taken  from  J;  so  borrow  3^ 

1  from  3.     1  equals  f ;  f  and  |  are  f ;  |  from  f  leaves  lf_ 

I;   |  — J.     2  from  8  leaves  1.  l^  Ans. 

7.  From      4^  subtract    2|.  H- 

8.  From       8-I  subtract     3f  4|. 

9.  From  23^  subtract  17^^  ^. 


UB 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


CASK    II, 

114.     When   the   fractions  have  not  a  common  denom- 
i  nator. 


1.  From  yV  subtract  |. 

Solution.— Ri^diicing  the  fractions  to  a 
common  denominator  (Art.  108),  f  =  ^J 
and  T%  =  H;   then,  f^  from    |-J  leaves  ^% 


OPKRATION. 


Explanation. — Since  the  denominators  are  different,  the  nu- 
merators do  not  express  parts  of  the  satne  size;  therefore,  one 
fraction  can  not  he  subtracted  from  the  other  till  they  are  reduced 
to  a  common  denominator. 

Bule. — 1.  Beduce  the  fractions  to  a  common  denomin- 
ator. 

2.  Fro7n  the  greater  numerator  subtract  the  less,  and  under 
the  remainder  icrite  the  common  denominator. 


2. 

From 

^  subtract 

i- 

3. 

From 

\  subtract 

i- 

4. 

From 

\  subtract 

1- 

5. 

From 

1  subtract 

h 

6. 

From 

1  subtracc 

A- 

7. 

From 

^  subtract 

1- 

8. 

From 

1^  subtract 

h 

9. 

From 

j-4^  subtract 

tV 

10. 

From 

^^  subtract 

T^- 

11. 

From 

3^  subtract 

If- 

.8 
15* 

fi- 
t's- 


tt- 


Solution. — \  equals  |,  and  |  equals  |.    |  can  Tiot       OPHUATloN^. 
be  taken  from  f ;   so  borrow  1   from  the  3.     1  equals  3| 

f ;  §  fi^nd  I  arc  |;  |  from  |  leaves  §.     2  from  3  leaves  1,  If 


12. 

From 

5 

subtract  |. 

13. 

From 

H 

subtract  4|. 

14. 

From 

"^1 

subtract  4f. 

15. 

From 

141 

subtract  12| 

16. 

From 

5A 

subtract  2i^. 

COMMON  FKAOTIo/s.  149 

If 

2H- 

17.  From  4^i^  subtract  3^.  4| 

18.  From  56^  subtract  421  14_i__. 

19.  From  60|  subtract  41^^.  19f 

20.  From  97|  subtract  48|.  48|. 


MULTIPLICATION   OF   FRACTIONS. 

115.     Multiplication    of  Fractions   is    the    process   of 
finding  the  product  of  two  or  more  fractional  numbers. 

1.  If  1  apple  cost  I  of  a  cent,  what  will  3  apples  cost? 

OPERATION. 

Solution. — They  will  cost  3  times  -|  of  a  cent  |  X  f  =  ¥' 

=  -L2_  of  a  cent  (Art.  101,  Prin.  1).     -i/  equals  y-  =  2f 


Explanation. — 3    apples    will    cost   | -j- | -f  |  =  ^^z.  of  a   cent; 
hence,  3  times  ^-=y,^-. 

2.  At  12  ct.  a  yard,  what  will  |  of  a  yard  of  ribbon 
cost? 

Solution. — \  of  a  yard  will  cost  \  of  12;=rij2  ^t.         operation. 
then,  I  of  a  yard  will  cost  2  times  -V— ^  ^^-  (  ^^^-  ^  )'      V"  X  f  =  "¥- 

¥  =  44.  ^  ^  .2^4^44 

3.  What  will  ^  of  a  yard  of  cloth  cost,  at  |  of  a  dollar 
pe**  yard? 


dollar;  then,  ^  of  a  yard  will  cost  4  times  ^3^  =  i|  of     |X  t=  35 
a  dollar. 


15a  RAY'S  NEA^  PHACTICAL  ARITHMETIC. 

Explanation. — f  of  I  of  a  dollar  is  -^^  of  a  dollar  (Art.  99); 
then,  }  of  I  of  a  dollar  is  3  times  -^^  =  /^  of  a  dollar  (Ex.  1). 

4.  MuUiply  f  by  f 

Solution. — J  is  the  same  as  ^  of  4  (Art.  96).  f  operation. 
multiplied  by  4  is  f  (Art.  101,  Prin.  1);  then,  f  multi-  f  n^  |_  ^8^ 
plied  hy  ^  of  4  is  J  of  1  =  ^5  (Ex.  3,  Explanation). 

Rule. — 1.  Multiply  together  the  numerators  of  the  given 
fractions  for  the  numerator  of  the  product. 

2.  Multiply  together  the  denominators  of  the  given  frac- 
tions for  the  denominator  of  the  product. 

Rem.  1. — Express  integers  in  the  form  of  fractions  (Art.  97,  3). 

Rem.  2. — Reduce  mixed  numbers  to  improper  fractions  (Art.  104). 
Sometimes  it  may  be  more  convenient  to  multiply  by  the  integral 
and  fractional  parts  separately. 

Rem.  3. — Indicate  the  operation  and  apply  the  Rule  for  Cancella- 
tion wherever  it  is  practicable  (Art.  91,  Rule). 

5.  Multiply  f  by  3.  2\ 

6.  Multiply  8  by  |.  ^ 

7.  Multiply  I  by  f.  ^, 

8.  Multiply  I  by  4.  2|, 

9.  Multiply  5  by  f .  3| 

10.  Multiply  f  by  |. 

operation. 

Solution. — Indicating  the  operation  and  ap-  ^ 

plying  the   Rule  for  Cancellation  (Art.  91),  the  |  X  f  — "  f 

result  is  §.  ^ 

11.  Multiply     I  by     6.  4. 

12.  Multiply  20  by     f.  15. 

13.  Multiply  ^\  by  if  H- 

14.  Multiply     I  by  10.  6. 


COMMON  FRACTIONS. 


151 


15.  Multiply  12  by  |. 

16.  Multiply  j\  by  f. 

17.  Multiply     f  by  6. 

18.  Multiply     7  by  |. 

19.  Multiply  21  by  3i. 


2JL 
91- 

To. 


Solution.  —  Reducing  2 J  and  SJ  to  im- 
proper fractions  (Art.  104),  they  are  |  and  |. 
Multiplying   together   |  and  |,  the  result  is 

6-3  ^:^  71^ 


OPERATION. 


--n 


20.  Multiply  18f  by  8. 


OPERATION. 


144  +  6  =  150. 


144 


150 


21.  Multiply  8  by  3|. 

22.  Multiply  2^  by  2|. 

23.  Multiply  10|  by  7. 

24.  Multiply  25  by  8|. 

25.  Multiply  -^%  by  17^. 

26.  Multii:)ly  lOf  by  9., 

27.  Multiply  64  by  8f. 

28.  Multiply  8f  by  f 


29i. 

H- 
7^. 
215. 
15^. 
97|. 
568. 

3f. 


Multiply  together : 


29. 
30. 
31. 
32. 
33. 
34. 
35. 


2A- 


TJ'     16' 

2-1-   -A-    li 

■^16'     IT'     *9- 

21. 


If 


6|,  2|, 

2h  3|,  4f, 

2i,  2A,  3i,.lT^. 

i     _3_      8.      5      ^      6 

8'     10?     9?    ^'     3'    T* 
1       9      4      7       .5       2      fi 

4'    T'    -^J     9^'     4'     3'    "• 


4091. 
49if 


22. 


152 


KAY'S  NEW  PRACTICAL  AKITHMETlC. 


36.  f  f,  If,  h  h  h  h  20. 

37.  2i,  6|,  ^,  ^3,  2,  f. 


24. 


116.     Fractioiuil 
multiplication. 


parts   of   integers    are    obtained    by 


1.  What  is  I  of  2? 

Solution.— J  of  2  is  f  (Art.  96);  then,  f  of  2  is 
2  times  ^  =  i.    i^^l 


OPERATION. 

i=n 

3f. 

2f 

8. 

10. 

12f 

18f 

Hi. 


117.     Compound    fractions    (Art.   99)    are    reduced    to 
simple  fractions  by  multiplication. 

1.  Reduce  |  of  ^  to  a  simple  fraction. 

Solution. — Multiplying  |  by  f  (Art.  115,  Rule),  operation. 

the  result  is  j%.  2y^*  =  j\ 


2. 

What  is 

f  of    5? 

3. 

What  is 

!  of    7? 

4. 

What  is 

4  of  10? 

5. 

What  is 

1  of  12? 

6. 

What  is 

1  of  15? 

7. 

What  is 

1  of  21  ? 

8. 

What  is 

-J^  of  25? 

9. 

What  is 

^  of  27? 

0. 

What  is 

^  of  28? 

2.  Reduce    ^  of  f  to  a  simple  fraction. 

3.  Reduce     f  of  |^  to  a  simple  fraction. 

4.  Reduce    i  of  |  of  2f  to  a  simple  fraction. 

5.  Reduce  ^^  of  f  to  a  simple  fraction. 

6.  Reduce     f  of  f  to  a  simple  fraction. 
I  of  4  of  1|^  to  a  simple  fraction. 


7.  Reduce 


3  3 

14 
3^ 
15 
■3T' 


COMMON   FRACTIONS.  153 

8.  Eeduce    f  of  f  of  i  to  a  simple  fraction. 

9.  Reduce     ^  of  |  of  f  to  a  simple  fraction.  2T 

10.  Bediice     f  of  |-  of  |^  to  a  simple  fraction. 

11.  Reduce     f  of  |  of  y7_  of  |f  to  a  simple  fraction. 

12.  Reduce  ^  of  |  of  |  to  a  simple  fraction. 

13.  Reduce  ^  of  |  of  1^  to  a  simple  fraction.  •^, 

14.  Reduce  f  of  ^  of  l\l  to  an  integer.  1 

15.  Reduce  f  of  2f  of  If  to  an  integer.  2, 

16.  Reduce  ^3  of  -/^  of  1|^  to  a  simple  fraction. 

17.  Reduce  ^  of  4  of  |  of  5  to  a  simple  fraction.  \, 

18.  Reduce  ^  of  |  of  f  of  |  of  f  of  |  of  j\  to  a  sim- 
ple fraction.  iV* 

Miscellaneous  Examples. 
118.     What  will  be  the  cost 

1.  Of  2^  lb.  of  meat,  at  l^  ct.  a  lb.?  30f  ct. 

2.  Of  3     yd.  linen,  at  $|  a  yd.  ?     Of  5  yd.?    Of  7  yd.? 
Of  61  yd.?     5fyd.?            '  S3f 

3.  Of  3^  lb.  of  rice,  at  4|-  ct.  a  lb.  ?  16  ct. 

4.  Of  3|  tons  of  iron,  at  $18f  per  T.?  $60. 

5.  Of  If  yd.  of  muslin,  at  $^\  per  yd.?  ${. 

6.  Of  21  lb.  of  tea,  at  $f  per  lb.?  $2. 

7.  Of  5|  cords  of  wood,  at  $1|  per  C?  $6|. 
'     8.  At  the  rate  of  b^  miles  an  hour,  how  far  will  a  man 

travel  in  7f  hours  ?  42|  mi. 

9.  I   own   I   of  a   steamboat,  and   sell   f  of  my  share : 
what  part  of  the  boat  do  I  sell  ?  f . 

10.  At   $6|   per  3^ard,  what  cost  |  of  a  piece  of  cloth 
containing  5^  yards?  $8^. 

11.  f  of  I  of  161-  X  I  of  I-  of  15  =:  what?  34f. 

12.  What  is  the  sum  of  f  +  J  and  |  X  I"?  l^' 


154  RAY'S  NEW  PRACTICAL  ARITHMETIC. 


DIVISION  OF   FRACTIONS. 

119.  Division  of  Fractions  is  the  process  of  finding 
the  quotient  of  two  fractional  numbers. 

1.  If  3  yards  of  ribbon  cost  f  of  a  dollar,  what  will  1 
yard  cost? 

OPERATION. 

Solution. — 1  yard  will  cost  J  of  f  =f  of  a  dollar      ^ 
(Art.  117).  ?Xi-f 

Explanation. — ^  is  to  be  divided  into  3  equal  parts.  Each  part 
will  be  2  (Art.  101,  Prin.  2);  for  f  :==  f -f- ^-f  ^. 

2.  At  2  dollars  a  yard,  what  part  of  a  yard  of  cloth 
can  be  bought  for  f  of  a  dollar? 

Solution. — For  1  dollar  \  a  yard  can  be  bought,       operation. 
and   for  ^  of  a  dollar  \  of   \^^^^  of  a  yard  (Art.       |X|  =  A 
117);   then,  for  |  of  a  dollar  3  times  -^^^-^^  of  a 
yard  can  be  bought. 

Explanation. — Were  it  required  to  find  how  many  yards,  at  $2 
a  yard,  could  be  bought  for  $6,  then  6  would  be  divided  by  2; 
hence,  to  find  the  part  of  a  yard  that  $|  will  pay  for,  \  must  be 
divided  by  2.  To  divide  \  by  2,  multiply  the  denominator  (Art. 
101,  Prin.  3). 

3.  At  I  of  a  cent  for  1  apple,  how  many  can  be  bought 
for  4  cents? 

Solution. — For  J  of  a  cent   \  an   apple  can  be  operation. 

bought,  and   for  |,  or  1   cent,  3   times  |  =  |  of  an  ^  • 

apple;  then,  for  4  cents,  there  can  be  bought  4  times  f  X  J  =  6 
I  ^=  6  apples. 

4.  At  f  of  a  cent  for  1  apple,  how  many  apples  can 
be  bought  for  f  of  a  cent? 


COMMON  FRACTIONS.  155 

of  a   cent  J  an  apple   can   be     operation. 
bought,  and    for  |,  or  1    cent,  3   times  i  =  |  of  an       |  X  1  =  f 
apple;  then,  for  \  of  a  cent  J  of  f  t=r  |  of  an  apple         1  =  1} 
can  be  bought  (Art.  117),  and  for  |  of  a  cent  3  times 
f  =  1-1-  apples. 

5.  Divide  |  by  4. 

Solution. — f   is   the   same   as   ^  of  4  (Art.  96).       operation. 
I  divided  by  4   is   j\  (Art.  101,  Prin.   3);    then,   f      sy^^  =  \^ 
divided  by  i  of  4Js  5  times  j\  =  i^  (Art.  115,  Ex.  1). 

Rule. — Mulflply  the  dividend  by  the  divisor  with  its  terms 
inverted. 


Rem.  1, — Express  integers  in  the  form  of  fractions  (Art.  97,  3). 
Rem.  2. — Reduce  mixed  numbers  to  improper  fractions  (Art.  104). 
Rem.  3. — Indicate  the  operation  and  apply  the  Rule  for  Cancel- 
lation whenever  it  is  practicable  (Art,  91,  Rule). 

6.  If  4  yards  of  muslin   cost  f  of  a  dollar,  what  will 
1  yard  cost?  Sf 

7.  At  i  a  cent  each,  how  many  apples  can  be  bought 
for  3  cents?  6. 

8.  At  ^  of  a  dollar  per  3^ard,  how  many  yards  of  mus- 
lin can  be  bought  for  Sy%  ?  ^' 

9.  If  1  orange   cost  3  cents,  what  part  of  an   orange 
could  be  purchased  for  ^  a  cent?  \. 

10.  At  f  of  a  dollar  per  yard,  how  many  yards  of  cloth 
can  you  buy  for  6  dollars?  8. 

11.  At  \  of  a   dollar   per   yard,  how   many  yards  of 
ribbon  can  be  purchased  for  f  of  a  dollar?  3f. 

12.  If  7  pounds  of  rice  cost  i|.  of  a  dollar,  what  will  1 
pound  cost?  $^. 


156 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


13.  Divide  4^-  by  If 

Solution. — Keducing  4^  and  IJ  to  improper  frac- 
tions (Art.  104),  we  have  |  and  |.  Dividing  |  by  J, 
the  result  is  31. 


14. 

Divide 

2|  by  6. 

15. 

Divide 

22  by  5J. 

16. 

Divide 

^  by  tV. 

17. 

Divide 

4f  by  8. 

18. 

Divide 

6     by  2f 

19. 

Divide 

4f  by  51. 

20. 

Divide 

124   by   11. 

21. 

Divide 

30  by  3f . 

22. 

Divide 

2\  by  7f 

23. 

Divide 

3|  by  7. 

24. 

Divide 

50  by  ^. 

25. 

Divide 

Iby  A. 

26. 

Divide 

47|  by  15. 

27. 

Divide 

56  by  5f 

28. 

Divide 

H  V  21. 

29. 

Divide  130|  by  18. 

30. 

Divide 

i  of  1  by  1 

off 


OPERATION. 

li=J 

Y  =  3| 

f- 

4. 

40. 

f- 

If- 
If 

8. 
^- 

1   I 

HA- 

25. 
lOf 

A- 


Explanation. — Invert  the  terms  of  both  operation. 

I  and  ^  as  in  the  case  of  the  divisor  being  a       jXfXIXi" 
simple  fraction. 


Of  17^. 


31.  Divide  f  of  f  by  f  of 

32.  Divide  i  of  5^  by  f 

33.  Divide  j\  of  |  of  12f^  by  ^  of  8|. 

34.  Divide  f  of  |  by  f  of  ^  of  5. 

35.  Divide  j\  of  f  of  12^3^  by  i  of  4^^  of  20. 


5 

6* 

i- 


120.     What  part  one    number  is  of  another  is    found 
by  division. 


COMMON  FRACTIONS. 

1.  1  is  what  part  of  2? 

Solution. — 1  is  i  of  2;  for  |  of  2  is  |,  or  1  (Art. 
98,  2d). 

2.  2  is  what  part  of  3? 

Solution. — 1  is  ^  of  3;    then,  2  is  2  times  1=1 
of  3. 

3.  ^  is  what  part  of  3? 
Solution. — 1  is  i  of  3;  then,  i  is  J  of  -J- 

4.  I  is  what  part  of  f  ? 

Solution. — J  is   i  of  f,  and  |,  or  1,  is  4  times 
=n:|  of  f;   then,  i  is   i   of  |  =  |  of  |,  and  |-  is  2 

imoQ    4  8    /-»f    3 


157 


i  of  3. 


t-l  of  f;  then, 
times  J  =  I  of  f . 


5.     3     is  what  part  of  4  ? 


6.     f     is 


what  part  of  5? 

7.  ^     is  what  part  of  i? 

8.  I    is  what  part  of  |? 

9.  3|  is  what  part  of  5  ? 

10.  I     is  what  part  of  f  ? 

11.  8|  is  what  part  of  11? 

12.  fi  is  what  part  of  |f? 


OPERATION. 


:Xi 


OPERATION. 

2  V  1 ^- 

1  /\   3  3 


OPERATION. 


OPERATION. 

2  \/   4  8 

3  A   3   —  ¥ 


2- 
4 
5^- 
3 

1  5 
16- 

I. 
9- 
_9 


121.     Complex    fractions,    (Art.    100)    are    reduced    to 
simple  fractions  by  division. 

1.  Eeduce   -|  to  a  simple  fraction. 

Solution.— Reducing  IJ  and  2i  to  improper  frac- 
tions (Art.  104),  we  have  {  and  |.  Dividing  |  by  J 
(Art.  119),  the  result  is  if. 


OPERATION. 


91  —  ^ 
^3  —  -i 


158  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  Eeduce   -^    to  a  simple  fraction.  ^. 

2 

3.  Keduce    -^    to  a  simple  fraction.  y2^. 

2 

4.  Keduce    —    to  a  simple  fraction.  ^. 

31 

5.  Reduce    j|    to  a  simple  fraction.  ^J|. 

21 

6.  Eeduce    -^    to  a  simple  fraction.  J|. 

7.  Eeduce    —    to  a  simple  fraction.  |. 

97 

8.  Eeduce  -r^    to  a  mixed  number.  44. 

8J 

9.  Eeduce    ^^    to  a  mixed  number.  If. 

^^ 
75 
10.  Eeduce  ^    to  a  simple  fraction.  |^. 


Miscellaneous  Examples. 

122.     1.    At  ^  a  dollar  per  yard,  how  many  yards  of 
silk  can  be  bought  for  $3^?  6^. 

2.  At  f  of  a  dollar  per  pound,  how  many  pounds    of 
tea  can  be  purchased  for  ^2^?  3|. 

3.  At   3|  dollars  per  yard    for  cloth,  how  many  yards 
can  be  purchased  with  $42^?  11^. 

4.  By   what   must   |    be    multiplied   that   the   product 
may  be  10?  26|. 

5.  Divide  3f  by  f  of  If  5f 

6.  Divide  ^  of  271  by  ^  of  21f  I29. 

7.  Multiply  li  by  i.  A- 


COMMON  FRACTIONS.  159 

8.  Multiply  JA  of  5^  by  ^.  |». 


'T2  ^T^  6 

li         2i 


9.  Divide  ^2  by  -f .  2 


10    Divide  —  bv  — ^. 


32- 


11. 


FRACTIONAL    COMPOUND    NUMBERS. 

128.     1.  Add  $16j\',  $9-1;  $53-V;  $2j|.  SSB^-V 

2.  I  paid  for  books   $9|-;    for  paper,  ^4^7^;    for  a  slate, 
$|;  for  pens,  $lf;  what  amount  did  I  expend?       S15^. 

3.  Having   $50^,   I  paid   a  bill   of  $27-^^:    how   much 
had  I  left?  S23Jg. 

4.  From  $32.31^  take  $15.12i.  $17.18f. 

5.  From  $5.81^  take  $1.18f.  ^4.621 

Find  the  cost  of 

6.  9  yd.  of  muslin,  at  121  ct.  a  yd.  S1.12f 

7.  21  lb.  of  sugar,  at  6^  ct.  a  lb.  $1.31f 

8.  15  yd.  of  cloth,  at  $3.18f  per  yd.  $47,811 

9.  51  yd.  of  linen,  at  $0,621  per  yd.  $3.43|. 

10.  121  yd.  of  ribbon,  at  18|  ct.  per  yd.  $2.34|. 

11.  131  yd.  of  calico,  at  16|  ct.  per  yd.  $2.25. 

12.  101  yd.  of  cloth,  at  $3,371  a  yard.  $34.59f. 

13.  17|  dozen  books,  at  $3.75  per  dozen.  $66.25. 

14.  At  18|  ct.  per  yard,  how  many  yards  of  muslin  can 
be  purchased  for  $2.25?     ,  12  yd. 

15.  At  371  ct.  per  bushel,  how  many  bushels  of  barley 
can  you  buy  for  $5.81|?  15^  bu. 

16.  If  5   yards   of  cloth   cost   $11.56^,  what  cost   one 
yard?  $2.31^. 

17.  Seven    men    share    $31.06^    equally:    what   is    the 
share  of  each  man?  $4.43f. 


IHO  KAY'S  NEW   PKACTR  AL  AKITHMETIC. 

18.  Eeduce  5  mi.  to  inches.  316800  in. 

19.  Keduee  2  mi.  2  rd.  2  ft.  to  feet.  10595  ft. 

20.  Reduce  20  yd.  to  rods.  ^  3  rd.  3|  yd. 

21.  Ecduce  15875  ft.  to  miles.  3  mi.  2  rd.  2  ft. 

22.  Reduce  U2634  in.  to  miles.      2  mi.  80  rd.  2  yd.  2  in. 

23.  How  many  steps,  of  2  ft.  8  in.  each,  will  a  man 
take  in  walking   2  miles?  3900. 

24.  How  many  revolutions  will  a  wheel,  of  9  ft.  2  in. 
circumference,  make  in  running  65  miles?  37440. 

25.  Reduce  1  A.  136  sq.  rd.  25  sq.  yd.  to  square  yards. 

8979  sq.  yd. 

26.  Reduce  7506  sq.  yd.  to  A.       1  A.  88  sq.  rd.  4  sq.  yd. 

27.  Reduce  5  chains  15  links  to  in.  4078|  in. 

28.  How  many  acres  in  a  field  40^  rd.  long  and  32  rd. 
wide?  8  A.  16  sq.  rd. 

29.  Reduce  4  years  to  hours.  35064  hr. 

30.  Reduce  914092  hr.  to  cen.     1  cen.  4  yr.  101  da.  4  hr, 

31.  In  what  time  will  a  body  move  fi'om  the  earth  to 
the  moon,  at  the  rate  of  31  miles  per  day,  the  distance 
being  238545  miles?  21  yr.  24|  da. 

124.  A  fraction  is  reduced  to  a  lower  denomination 
by  multiplication  (Art.  63,  Rule  I). 

1.  Reduce  J^  of  a  peck  to  the  fraction  of  a  i)int. 

Solution.  —  To   reduce    .}^  of  a  jx'ck  to  the        opkration. 
fraction  of  u  pint,  multiply  by  8  unci  by  2.     The     ^  X  I  X  r  =-  I 
result  is  |  of  a  pint.  ^ 

2.  Reduce  -^^  bu.  to  the  fraction  of  a  quart.  |. 

3.  Reduce  -^^  lb.  to  the  fraction  of  an  ounce.  ^. 

4.  Reduce  yL.  lb.  Troy  to  the  fraction  of  an  ounce.  |. 

5.  Reduce  ^  rd.  to  the  fraction  of  a  foot.  |^. 

6.  Reduce  y^Vo"  "^'  ^^  ^^^^  fraction  of  a  square  rod.  |. 


COMMON  FK ACTIONS.  161 

7.  Reduce  $3!^  to  the  fraction  of  a  cent.  f. 

8.  Eeduce  y^^g^  da.  to  the  fraction  of  a  minute.        |^. 

9.  Reduce  -^^  bu.  to  the  fraction  of  a  pint.  |. 

125.  In  reducing  a  fraction  to  a  lower  denomination, 
when  the  result  is  a  mixed  number,  proceed  only  with 
the  reduction  of  the  fractional  part.  This  is  called  find- 
ing the  value  of  a  fraction  in  integers. 

1.  Find  the  value  of  f  of  a  day  in  integers. 

Solution. — To  reduce  |  of  a  day  to  hours,  mul- 
tiply by  24;  the  result  is  9|  hr.  To  reduce  |  of  an 
hour  to  minutes,  multiply  by  60;  the  result  is  30 
min.     I  of  a  day,  then,  is  9  hr.  36  min. 

2.  Find  the  value  of  4  mi.  in  integers. 

3.  Find  the  value  of  $|-  in  integers. 

4.  Find  the  value  of  f  mi.  in  integers. 

5.  Find  the  value  of  ^  lb.  Troy  in  integers. 

9  oz.  12  pwt. 

6.  Find  the  value  of  -^^  T.  in  integers.        8  cwt.  75  lb. 

7.  Find  the  value  of  |  A.  in  integers.  100  sq.  rd. 

8.  Find  the  value  of  -J  of  63  gallons  of  wine  in  in- 
tegers. 55  gal.  1  pt. 

12G.  A  fraction  is  reduced  to  a  higher  denomination 
by  division     (Art.  63,  Rule  IT). 

1.  Reduce  |  of  a  pint  to  the  fraction  of  a  peck. 

Solution.— To  reduce  |  of  a  pint  to  the  frac-  operation. 

tion  of  a  peck,  divide  by  2  and  by  8.     The  result     |  X  ^  X  i  =  2? 
is  2V  of  a  peck. 


operation. 

-I  X  -\^-  -^  9f 

|X¥  =  36 

256  rd. 

60  ct. 

128  rd. 

2.  Reduce  4  qt.  to  the  fraction  of  a  bushel.  ^. 

luce  4 

Prac.  11. 


3.  Reduce  4  ft.  to  the  fraction  of  a  rod.  y 


162  RAY'S  NEW  PKACTICAL  ARITHMETIC. 

4.  Eeduce  ^\  oz.  to  the  fraction  of  a  pound.  rAir- 

5.  Keduce  ^  lb.  to  the  fraction  of  a  ton.  t^Vtt- 

6.  Eeduce  f  pt.  to  the  fraction  of  a  bushel.  -^^. 

7.  Eeduce  ^  oz.  to  the  fraction  of  a  hundred-weight. 

2800- 

8.  Eeduce  f  in.  to  the  fraction  of  a  rod.  ^J^ 

9.  Eeduce  |  min.  to  the  fraction  of  a  day.  rwru- 
10.  Eeduce  yf^  ^^-  to  the  fraction  of  a  hundred-weight. 

127.  To  find  what  part  one  compound  number  is  of 
another,  reduce  them  to  the  same  denomination  and  pro- 
ceed as  in  Art.  120. 

1.  2  ft.  3  in.  is  what  part  of  a  yard? 

OPERATION. 

Solution. — 2  ft.  3  in.  equals  27  in.     I  yd.        2  ft.  3  in.  =  27  in. 
equals  36  in.     27  in.  are  |J  of  36   in.     fj  1  yd.=r=36in. 

equals  j.     2  ft.  3  in.,  then,  is  }  of  a  yard.  ii  =  i 

2.  2  ft.  6  in.  is  what  part  of  6  ft.  8  in.?  | 

3.  2  pk.  4  qt.  is  what  part  of  a  bushel  ?  | 

4.  What  part  is  2  yd.  9  in.  of  8  yd.  2  ft.  3  in.  ?  ^ 

5.  What  part  of  a  day  is  13  hr.  30  min.  ?  ^ 

6.  What  part  of  a  mile  is  145  rd.?  fj 

7.  What  part  of  a  yard  is  2  ft.  8  in.?  « 

8.  15  mi.  123  rd.  is  what  part  of  35  mi.  287  rd.?        f. 

9.  A  man  has  a  farm  of  168  A.  28  sq.  rd. ;  if  he  sell 
37  A.  94  sq.  rd.,  what  part  of  his   farm   will    he   dispose 

of?  A\- 

10.  What  part  of  a  pound  is  7^  oz.  ?  ^. 

11.  2  qt.  lipt.  is  Avhat  part  of  1   bu.   1  qt.  If  pt.  ? 

1  6 

12.  1  yd.  1  ft.  1^  in.  is  what   part   of  3   yd.  2  ft.  8f 

in   ?  1 91 « 

1^-  •  5  4T3' 


COMMON  FKACTIONS. 


163 


128.  To  add  and  subtract  fractional  compound  num- 
bers, find  the  value  of  the  fractions  in  integers  and  then 
proceed  as  in  Addition  and  Subtraction  of  Compound 
Numbers. 


1.  Add  I  yd.  and  f  ft. 

Solution. — -|  yd.  equals  2  ft.  3  in.;  |  ft. 
equals  10  in.;  the  sum  of  2  ft.  3  in.  and  10 
in.  is  3  ft.  1  in.  (Art.  75). 


OPERATION. 

yd.  :rrr2  ft.  3  in. 

ft.    =         10  in. 

3  ft.  1  in. 


2.  From  |  da.  subtract  |  hr. 


Solution. — |  da.  equals  5  hr.  20  min.; 
I  hr.  equals  50  min.;  50  min.  subtracted 
from  5  hr.  20  min.  leaves  4  hr.  30  min. 
(Art.  76). 


OPERATION. 

I  da.  ==5  hr.  20  min. 

I  hr.  =  50  min. 

4  hr.  30  min. 


3.  Add  I  da.  and  f  hr.  16  hr.  45  min 

4.  Add  \  wk.  i  da.  and  \  hr.  2  da.  15  min 

5.  Add  I  wk.  I  da.  |  hr.  and  |  min. 

5  da.  6  hr.  40  sec 

6.  Add  \^  gal.  and  ^2  ^t. 

7.  From  ^  da.  subtract  Jg  hr. 

8.  From  H  subtract  SA. 


3  qt.  1  pt.  2  gi. 

18  hr.  36  min.  40  sec. 

55  ct. 


9.  From  |  lb.  subtract  ^  oz. 


^oz. 


10.  From  |  da.  subtract  ^  hr.         2  hr.  34  min.  17|  sec. 


129.     Promiscuous   Examples. 


1.  Reduce  y  g  |  j  to  its  lowest  terms. 
2     Arid     -^       8     91    ^2 

3.  From  34^  subtract  1^. 

4.  From  3|  subtract  ^  of  3^. 

5.  Add  f  of  j\  and  |  of  -j^. 


a- 


16^  RAY'S  NEW  PRACTICAL  xVRITIlMETIC. 

G.  Add  1|  -^  2.1-  and  ^  -^  3i  2f|. 

7.  What  niiinber  divided  by  f  will  give  10  for  a  quo- 
tient? 6. 

8.  What  number  multiplied  by  |  will  give  10  for  a 
product?  16|. 

9.  What  number  is  that,  from  which  if  you  take  ^  of 
itself,  the  remainder  will  be  16?  28. 

10.  What  number  is  that,  to  which  if  you  add  j^  of 
itself,  the  sum  will  be  20?  14. 

11.  A  boat  is  worth  S900 ;  a  merchant  owns  |  of  it,  and 
sells  ^  of  his  share :  what  part  has  he  left,  and  what  is  it 
worth?  *  3^  left,  worth  $375. 

12.  I  own  j^  of  a  ship,  and  sell  ^  of  my  share  for 
$1944|:  what  is  the  whole  ship  worth?  810000. 

13.  What  part  of  3  cents  is  |  of  2  cents?  |. 

14.  What  part  of  368  is  170?  i|. 

15.  From  |^  subtract  the  sum  of  ^,  y^y,  and  Jj^y. 

1  007 

16.  From  1  subtract  -^^  of  ^^  of  4yV  y\. 

17.  From  |  ^  f  subtract  |  -^  |f  •  2%- 

18.  If  I  ride  2044  rods  in  y^^  of  an  hour,  at  that  rate 
how  far  will  I  ride  in  1|4  hr.  ?  8468  rd. 

19.  What  part  of  1^  feet  are  3^  inches?  |. 

20.  Two  men  bought  a  barrel  of  flour ;  one  paid  S3^, 
and  the  other  $3| :  what  part  of  it  should  each  have  ? 

One  ^^^,  the  other  y^^^. 

21.  A  has  $2400  ;  |  of  his  money,  +  $500,  is  |  of  B's  : 
what  sum  has  B?  $1600. 

22.  John  Jones  divided  his  estate  among  2  sons  and  3 
daughters,  the  latter  sharing  equally  Avith  each  other. 
The  younger  son  received  $2200,  which  was  -^  of  the 
share  of  the  elder,  w^hose  share  was  ^f  of  the  whole 
estate :  find  the  share  of  each  daughter.  $1356^. 


130.    An  aliquot  part  is  an  exact  divisor  of  a  number. 
Aliquot  Parts  of  100. 


5    =iV 

124  =i 

25    =i 

6i=TV 

ie|  =  * 

33J  =  i 

"    =tV 

20    =1- 

50    =1 

The  following  multiples  of  aliquot  parts  of  100  are  often 
used  ;  184=^3^,  37i=a,  40=f,  60=f,  624=t,  75=f,  87*=^ 

1.  What  will  24  yd.  of  muslin  cost  at  25  ct.  a  yd.  ? 

OPERATION. 

Solution. — Since  25  ct.  is  \  of  a  dollar,  the  cost  will  4)24 

be  ^  as  many  dollars  as  there  are  yards.    \  of  $24  is  $6.  $  6 

2.  I   spent   $1,121  for  muslin    at   12|   ct.    a    yd. :    how 
many  yd.  did  I  bii}  ? 

OPERATION. 

Solution. — Since  V2h  ct.  is  I  of  a  dollar,  there  will  1"^ 

he  8  times  as  many  yards  as  there  are  dollars.     8  times  8 

H  =  9yd.  9"  yd. 

8.  What  cost  12^-  yd.  of  ribbon  at  ]8|  ct.  a  yd?     S2.34g. 


166  KAY'S  NEW  PKACTICAL  ARITHMETIC. 

4.  Paid  $2.25  for  muslin  at  18J  ct.  a  yd.  how  many  yd. 
did  I  buy?  12  yd. 

5.  What  will  5^  yd.  of  linen  cost  at  S0.62i  a  yd.? 

$3.43f. 

6.  Paid  $66.25   for  books  at  $3.75  a  dozen :   how  many 
doz.  books  did  I  buy?  17|  doz. 

7.  What  will  80  gal.  of  wine  cost  at  $2.37^  a  gal.? 

$190. 

8.  A  number  of  men  divide  $39  so  that  each  one  re- 
ceives $4.87^:  how  many  men  are  there?  8. 

9.  What  will  36  barrels  of  flour  cost  at  $8.33^  a  bar- 
rel ?  $300. 

10.  How   many  yd.  of  cloth   at    $1.33^  a   yd.   can   be 
bought  for  $246. 66|?  185  yd. 

11.  What  will   4  A.   60  sq.  rd.  of  land  cost  at    $16.50 
an  acre? 

Solution. — Since    1   A.   costs  $1G.50,  operation. 

4  A.  cost   $16.50X4:=  $66.     Since   160  $16.50 

sq.  rd.  =  1   A.,  40  sq.  rd.  =  \  A.     The  ^ 

cost  of  40  sq.  rd.  will  be  J  of  $16.50=  66.00 

$4,121.     The  cost  of  20  sq.  rd.  will  be  J  J  of  $16.50   =   4.12^ 

of  the  cost  of  40  sq.  rd.,  or  $2.06^.     The  J  of      4.12^  =   2mI 
total    cost    is    $66 -f  $4.121 -f  $2.06}=  $72.18J 

$72.18|. 

12.  At  $18.33^  per  acre,  how  much  land  can  be  bought 
for  $229,162?  mA. 

13.  What    will    11    A.    120    sq.    rd.    of   land    cost,    at 
$125.60    per    acre?  $1475.80. 

14.  At  $250  a  lot,  containing  50  X  150    ft.,  how    much 
land  can  be  bought  for  $10000? 

6  A.  141  sq.  rd.  28  sq.  jd.  108  sq.  in. 

15.  What  will  83  bu.  3  pk.  2  qt.  of  grass  seed  cost,  at 
$6.20  a  bu.?  $519.63|. 


PKACTICE.  167 

16.  At  $0.75  a  bushel,  how  raany  bushels  can  be 
bought   for   S167.50?  223  bu.  1  pk.  2  qt.  li  pt. 

17.  What  will  3|  yd.  cost,  at  $1.75  a  yard?         $6.12f 

18.  At  SI. 50  a  yard,  how  much  cloth  can  be  bought 
for    S7.12I-?  4f  yd. 

19.  What  will  45  lb.  12  oz.  of  butter  cost,  at  $0,371 
per  pound?  $17.15|. 

20.  At  $0,121  per  pound,  how  much  sugar  can  be 
bought  for  $2.93f  ?  23-i   lb. 

21.  What  is  the  cost  of  2  T.  9  cwt.  of  wool  at  37^  ct. 
a  pound?  $1837.50. 

22.  What  is  the  cost  of  100  readers  at  $3.90  a  dozen? 

$32.50. 

23.  What  is  the  cost  of  3f  dozen  knives  at  $5.40  a 
dozen?  $20.25. 

24.  A  farmer  sold  6^  doz.  chickens,  at  $0,331  apiece, 
and  37^  lb.  butter,  at  $0.37^  per  pound :  he  received  $36 
in  money,  and  the  remainder  in  sugar,  at  $0.12^  per 
pound:  how  many  pounds  of  sugar  did  he  get?       32i  lb. 


131.     The    orders    of   integers    decrease    from    left    to 
right  in  a  tenfold  ratio. 

Thus,  in  the  mmibcr  1111,  1   thousand  is  10  times  1  hundred,! 
hundred  is  10  times  1  ten.  and  1  ten  is  10  times  1  unit. 


ORDERS   OP    DECIMALS. 

132.  1.  The  orders  may  be  continued  from  the  order 
units  toward  the  right  by  the  same  law  of  decrease. 

2.  Let  tlie  order  units  be  separated  from  the  order 
that  follows  by  a  point  (.). 

3.  Then,  in  the  number  1.111, 

Ist.  Since  the  1  to  the  left  of  the  point  is  1  unit,  the  1  to  the  riu;ht 
of  the  point  is  1  tenth;  for  1  unit  is  10  times  J^. 

2d.  Since  the  first  order  from  the  unit  is  1  tenth,  the  second  order 
from  the  unit  is  1  hundredth;    for  ^^  is  10  times  yi^. 

3d.  Since  the  second  order  from  the  unit  is  1  hundredth,  the  third 
order  from  the  unit  is  1  thousandth;  for  yi^  is  10  times  yoVo- 

4th,  In  like  manner  it  may  be  shown  that  1  in  the  fourth  <n"der  to 
the  right  from  the  unit  is  1  ten-thousandth;  1  in  the  fifth  order  to 
the  riglit  is  1  hundred-thousandth;  1  in  the  sixth  order  is  1  millionth, 
etc. 

Rem. — A  number  consisting  of  figures  other  than  1,  might  be  used 
as  well  for  the  purpose  of  illustration, 
(1G8) 


DECIMAL  FRACTIONS.  169 

4.    The    position    of  the    integral    and    decimal    phices 
relative  to  the  unit  ia  exhibited  in  the  Ibllowinii; 


DTACiRAM. 

1       1.       1 


^^y^. 


\A 


/-     .^c)V^ 


5.  The  first  order  on  the  left  of  the  unit  is  tens,  the 
first  order  on  the  right  of  the  unit  is  tenths;  the  second 
order  on  the  left  is  hundreds;  the  second  order  on  the 
right  is  hundredths,  etc. 


DEFINITIONS. 

133,  1.  A  decimal  fraction,  or  decimal,  is  one  or  more 
tenths,  hundredths,  thousandths,  etc.,  written  like  the  orders 
of  integers. 

2.  A  decimal  point  ( . )  is  placed  before  the  order 
tenths  to  distinguish  the  fraction. 

3.  The  decimal  orders  increase  from  right  to  left,  and 
decrease  from  left  to  right  the  same  as  the  orders  of 
integers. 

4.  The  names  of  the  orders  of  decimals  are  similar  to 
the  names  of  the  corresponding  orders  of  integers. 

134.  Conversion  of  the  common  fractions  ■^,  j^-^, 
iiro¥?  ^^^'■•>  ^^  decimals. 


170 
1. 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


yi^    is    written  .1 


j\  are  written  .2 


4 
TIT 

5 
TIT 


.4 

.5 


A 

are 

written 

.6 

A 

u 

-' 

.7 

8 

u 

u 

.8 

T% 

u 

u 

.9 

Hence,  i^*/ien  the  denominator  is  10,  fAere  is  one  decimal 
order. 

2.     Yw^  ^^  wi-itten  .01;  there  being  no  tenths,  a  cipher 
is  written  in  the  vacant  order. 


yf^  are  written  .02 

j^^  are  written  .06 

yfir    "          "        -03 

T^     "          "        -07 

T^T     "          "        •O't 

Ths    "         "        -08 

t4i7    "         '•        -05 

T^tj    "          "        -09 

.Hence,  when  the  denominator  is  100,  there  are  tiro  deci- 
mal orders. 

3.     yo^oTj  is    written    .001 ;  there   being    no  tenths   and 
no  hundredths,  ciphers  are  written  in  the  vacant  orders. 


lOOlF 
3 

5 

Tirxro 


.003 
.004 
.005 


Att  ^i*e  written  .006 


T7T0¥ 
8 

9 
TOGO 


.007 
.008 
.009 


Hence,   when    the   denominator   is  1000,   there   are   three 
decimal  orders. 


4.  In  like  manner; 


1  OUTFO 


is     written 


TTr¥77"00^ 


.0001 

.00001 

.000001 


DECIMAL  FRACTIONS.  171 

Hence,  the  number  of  orders  in  the  decimal  is  always 
the  same  as  the  number  of  ciphers  in  the  denominator  of 
the  common  fractio7i. 

5.  j\  and  j^-^  are  jW  written         .11 

1  1  1  1  f\V(^      1111  '^  1111 

Hence,  tenths  and  hundredths  are  read  as  hundredths; 
tenths,  hundredths,  and  thousandths  are  read  as  thou- 
sandths;  tenths,  hundredths,  thousandths,  and  ten-thou- 
sandths are  read  as  ten-thousandths,  etc. 

6.  The  numerator  of  a  decimal  is  the  number  it  ex- 
presses disregarding  the  decimal  point. 

7.  If  there  are  vacant  orders  before  the  numerator, 
ciphers  are  written  in  them. 

8.  The  name  of  the  right  hand  order  is  the  name  of 
the  decimal. 

To  Wj^ite  Decimals. 

135.     1.  Write  two  hundred  and  sixty-five  thousandths. 

Number  Written.         .265. 

Explanation. — First,  write  the  numerator,  265,  as  an  integer. 
The  figure  5  must  stand  in  the  order  thousandths  (134,  8);   then, 

6  must  be  hundredths  and  2   must  be  tenths;    the  decimal   point, 
therefore,  is  phiced  before  the  figure  2  (133,  2). 

2.  Write  two  hundred  and  sixty-five  millionths. 

Number  Written.        .000265. 

Explanation. — Write  the  numerator,  265,  as  an  integer.  The 
figure  5  must  stand  in  the  order  millionths  (134,  8);  then,  6  must  be 
hundred-thousandths,  2  must  be  ten-thousandths,  and  ciphers  must 
be  written  in  the  orders  thousandths,  hundredths,  and  tenths  (134, 

7  );  the  decimal  point  is  placed  before  0  tenths  ( 133,  2 ). 


172  KAY'8  NEW  PRACTICAL  ARITHMETIC. 

3.  Write  two  huiidrod  and  Hixty-fivc  hundredths. 

NuMJJKR  Written.        2.65. 

Explanation. — Write  the  numerator,  2G5,  as  an  integer.  The 
figure  5  must  stand  in  the  order  hundredths;  then,  0  must  be  tenths; 
the  decimal  point,  therefore,  is  placed  between  the  figures  2  and  6. 

4.  Write  four  huiidred  and  ninety-eight  and  two  hun- 
dred and  sixty-five  miUlonths. 

NuMBKR  Written.        498.000265. 

Explanation. — First  write  the  decimal  as  in  Ex.  2;  then  write 
the  integer,  placing  it  at  the  left  of  the  decimal  point. 

Rule. — 1.    Write  the  numerator  as  an  integer. 
2.  Place  the  decimal  point  so  that  the  name  of  the  right 
hand  order  shall  be  the  same  as  the  name  of  the  decimal. 

Note. — Pupils  should  be  rendered  familiar  with  the  decimal 
orders  so  as  to  name  them  readily,  in  supcession,  both  from  loft  to 
right,  and  from  right  to  left. 

Rem.  1. — When  the  decimal  is  a  proper  fraction  it  is  sometimes 
necessary  to  prefix  ciphers  to  the  numerator  ( Ex.  2  ). 

Rem.  2. — When  the  decimal  is  an  improper  fraction,  the  decimal 
point  is  placed  between  two  of  the  figures  of  the  numerator  (  Ex.  3). 

Rem.  3. — In  a  mixed  number,  the  decimal  point  is  placed  after  the 
units  order  of  the  integer  (  Ex.  4  ). 

Write  the  following  decimal  numbers: 

5.  Twenty-six  hundredths. 

G.  Thirty-five  himdredths. 

7.  Eighty-seven  hundredths. 

8.  Four  hundred  and  nineteen  hundredths. 

9.  Five  thousandths. 

10.  Fifty-four  thousandths. 

11.  Three  hundred  and  four  thousandths. 


DECIMAL  FK ACTIONS.  173 

12.  Seven    thousand    two     hundred    and    ninety -three 
thousandths. 

13.  Twenty-five  and  forty -seven  thousandths. 

14.  Tw^o  hundred  and  five  ten-thousandths. 

15.  Four   thousand    one    hundred  and    twenty -five  ten- 
thousandths. 

16.  Mne  hundred -thousandths. 

17.  Nine  hundred  thousandths. 

18.  Six  hundred  and  five  hundred-thousandths. 

19.  Twenty  thousand  three  hundred  and  four  hundred- 
thousandths. 

20.  Seven  millionths. 

21.  Two  hundred  and  three  millionths. 

22.  Three  hundred  thousand  and  four  millionths. 

23.  Twenty-four  ten-millionths. 

24.  Eighty  thousand  and  six  ten-millionths. 

25.  Two  hundred  millionths. 

26.  Two  hundred-millionths. 

27.  Nine  hundred  and  seven  hundred-millionths. 

28.  Twenty  million    twenty  thousand    and    three    hun- 
dred-millionths. 

29.  One    million    ten    thousand    and  one  hundred  mill- 
ionths. 

30.  One    million   ten    thousand    and    one  hundred-mill- 
ionths. 

31.  One  hundred  and  six  and  thirty-seven  thousandths. 

32.  One  thousand  and  one  thousandth. 

33.  Two  hundred  and  twenty-five  thousandths. 

34.  Two  hundred  units  and  twenty-five  thousandths. 

35.  Two    thousand    nine     hundred     and     twenty-nine 
millionths. 

36.  Two    thousand    nine    hundred    units    and    twenty- 
nine  millionths. 

37.  One  million  and  five  hillionths. 


174  HAY'S  NEW  PRACTICAL  ARITHMETIC. 

38.  Two  hundred  and  two  ten-biUionths. 

39.  Two  hundred  units  and  two  ten-hillionths. 

40.  Sixty-five  and  six  thousand  and  five  inilUonths. 

Change    the    following    common    fractions  to  decimals: 

4.1         3  7  9  17  2  3  4  1  5  3 

'^^^  tW'    tVh'    iWtt'    tWttj    AVtf'    tVV^- 

A*:{  3  10  1  5  3  5  0  3 


To  Read  Decimals. 
136.     1.  Read  .2G5. 

Number  Read. — Two  hundred  and  sixty-five  thousandths. 

Explanation. — Disregarding  the  decimal  point,  the  number  is 
two  hundred  and  sixty-five;  this  is  the  numerator  of  the  decimal 
(134,  6).  The  right  hand  order  of  the  decimal  is  thousandths;  this 
is  the  name  of  the  decimal  (134,  8). 

2.  Read  .000265. 

Number  Read. — Two  hundred  and  sixty-five  millionths. 

Explanation. — Disregarding  the  decimal  point,  the  number  is 
two  hundred  and  sixty-five;  this  is  the  numerator  of  the  decimal. 
The  right  hand  order  is  millionths;   this  is  the  name  of  the  decimal. 

3.  Read  2.65. 

Number  Read. — Two  and  sixty-five  hundredths,  or  two  hundred 
and  sixty  five  hundredths. 

Rule. — 1.  Disregarding  the  decimal  pointy  read  the  num- 
ber as  an  integer. 

2.   Give  the  name  of  the  right  hand  order. 


DECIMAL  FRACTIONS.  175 

Note. — Before  commencing  to  read  the  decimal,  the  name  of  the 
right  hand  order  should  be  ascertained  (135,  Note,  under  Rule). 

Rem. — A  mixed  number  may  be  read  either  as  an  integer  and  a 
fraction,  or  as  an  improper  fraction  (Ex.  3). 

Eead  the  following  decimal  numbers : 

4.  .028;     .341;     2.327;     50.005;     184.173. 

5.  .0003;     .0625;     .2374;     .2006;     .0104. 

6.  3.0205;     810.2406;     10720.0905. 

7.  .00004;     .00137;     .02376;     .01007. 

8.  .001768;     .040035;     70.360004. 

9.  .1010101;     .00040005;     .00100304. 

10.  .31456;     .000133;     60.04;     45.1003. 

11.  357.75;     .4928;     5.945;     681.0002. 

12.  70.1200764;     954.203;     38.027. 

13.  1007.3154;  7496.35491768. 

14.  .00715;  3.00005;  28.10065701. 

15.  13.0008241094710947. 

Change  the  following  decimals  to  common  fractions, 

16.  .9;     .13;     .19;     .29;     .37;     .73. 

17.  .91;     .347;     .513;     .691;     .851;     .917. 

18.  .007;     .0207;     .00079;     .001007. 

19.  1.36;     .3421;     .03401;     .0900. 

20.  .001;     .5302;     8.01;     .000053. 

137,  The  operations  with  decimals  are  Reduction^  Ad- 
dition, Subtraction,  Multiplication  and  Division. 

REDUCTION   OF  DECIMALS. 

138.  Keduction  of  Decimals  is   changing  their  form 
without  altering  their  value.     There  are  four  cases. 


176  KAY'S  NEW  PRACTICAL  ARITHMETIC. 


CASE    I. 

139.  1.  Annexing  decimal  ciphers  to  an  integer  does  not 
change  its  vdlue. 

Thus,  7.00  is  the  sume  as  7;  for  7.00  is  7  and  no  hundredths  (Art. 
136,   Rule). 

2.  Conversely :    Omitting  decimal   ciphers  from   the  right 
of  an  integer  does  not  change  its  value. 

Number  1  of  this  case  evidently  corresponds  to  Case  I,  Art.  103, 
and  2  to  Case  III,  Art.  105. 

CASE    II. 

140.  1.    Annexing  ciphers   to  a.  decimal  does  not  change 
its  value. 

Thus,  .70  is  the  same  as  .7;  for  y^  — y^o^. 

2.  Conversely:     Omitting   ciphers   from   the  right   of  a 
decimal  does  not  change  its  value. 

Number  1  of  this  case  evidently  corresponds  to  case  IV,  Art.  106, 
and  2  to  Case  V,  Art.  107. 

CASE   III. 

141.  To  reduce  a  decimal  to  a  common  fraction. 

1.  Eeduce  .75  to  a  common  fraction. 

Solution. — 75   hundredths  written    as  a  common        operatiox. 
fraction  is  j^j^^.     -^^^  reduced  to  its  lowest  terms  (Art.  .75  =  -^^-^ 

107),isf.  i%=l 

Bule. — 1.    Write  the  decimal  as  a  common  fraction. 

2.  Reduce  the  fraction  to  its  lowest  terms. 


T6"- 


JL9_ 
12  8- 


^5U- 


DECIMAL  FKACTIONS.  177 

2.  Reduce       .6  to  a  common  fraction.  |. 

8.  Reduce     .25  to  a  common  fraction.  \. 

4.  Reduce  .375  to  a  common  fraction.  f. 

5.  Reduce  .035  to  a  common  fraction.  2-^^. 

6.  Reduce  .5625  to  a  common  fraction. 

7.  Reduce  .34375  to  a  common  fraction.  \ 

8.  Reduce  .1484375  to  a  common  fraction. 

9.  Express  4.02  as  an  integer  and  common  fraction 

4 
10.  Express  8.415  as  an  integer  and  common  fraction 

CASE    IV. 

142.     To  reduce  a  common  fraction  to  a  decimal. 

1.  Reduce  f  to  a  decimal. 

Solution. — Annexing  a  decimal  cipher  to  3,  it  is  operation. 
3.0;  30  tenths  divided  by  4  is  7  tenths,  and  2  tenths  4 )  3.00 

remaining.     Annexing  a  cipher  to  .2   it   is  .20;    20  .7  5 

hundredths    divided    by   4    is    5    hundredths.      The 
result   is   .75. 

Explanation. — J  is  3  divided  by  4  (Art.  97 ).  Annexing  a 
decimal  cipher  to  3  does  not  change  its  value  (Art.  139).  Annexing 
a  cipher  to  .2  does  not  change  its  value  (Art.  140). 

B.ule. — 1.  Annex  decimal  ciphers  to  the  numerator. 

2.  Divide  by  the  denominator. 

3.  Point  off  as  many  decimal  orders  iyi  the  quotient  as 
there  are  decimal  ciphers  annexed  to  the  numerator. 

2.  Reduce         I  to  a  decimal.  .8 

3.  Reduce         f  to  a  decimal.  .625 

4.  Reduce       2V  ^^  ^  decimal.  .28 

Prac.  12. 


178  KAY'S  NEW  PRACTICAL  AKITHMETIC. 

5.  Eeduce       -^jj  to  a  decimal.  .075 

6.  Eeduce       ^|  to  a  decimal.  .9375 

7.  Eeduce  yrsir  *^  *^  decimal.  .0008 

8.  Eeduce     -^^j^  to  a  decimal.  .0225 

9.  Eeduce     ^l^  to  a  decimal.  .00390625 

10.  Eeduce         |^  to  a  decimal.  .83  -\- 

11.  Eeduce       ^  to  a  decimal.  .09  + 

12.  Eeduce       ^\  to  a  decimal.  .12 -|- 


ADDITION   OF   DECIMALS. 

143.     Addition  of  Decimals  is  the  process  of  finding 
the  sum  of  two  or  more  decimal  numbers. 

1.  Add  375.83;  49.627;  5842.1963;  813.9762. 

Solution. — Write    the    numbers   so  that   the  operation. 

four  decimal  points  may  be   in    a    column,   the  3  7  5.83 

units  5,  9,  2,  3  in  the  first  cohimn  to  the  left,  the  4  9.62  7 

tenths  8,  6,  1,  9  in  the  first  column  to  the  right,  5  842.196  3 

etc.;  then,  adding  as  in  simple  numbers,  place  the  813.9762 

decimal  point  in  the  sum  between  1  and  6  under  7081.6295 
the  column  of  decimal  points. 


Rule. — 1.  Write  the  ninnhers  so  that  the  decimal  points 
and  figures  of  the  same  order  may  stand  in  the  same 
cohunn. 

2.  Add  as  in  simple  numbers. 

3.  Place  the  decimal  point  in  the  sum  under  the  column 
of  decimal  points. 

2.  Add  37.1065;  432.07;  4.20733;  11.706.         485.08983 

3.  Add  4  and  4  ten-thousandths;  28  and  35  thou- 
sandths; 8  and  7  hundredths;  and  9404  hundred-thou- 
sandths. 40.19944 


DECIMAL  FEACTIOISIS.  179 

4.  Find  the  sum  of  3  units  and  25  hundredths;  6 
units  and  4  tenths;  and  35  hundredths.  10. 

5.  Add  21.611;  6888.32;  3.4167.  6913.3477 

6.  Add  6.61;  636.1;  6516.14;  67.1234;  and  5.1233. 

7231.0967 

7.  Add  4  and  8  tenths ;  43  and  31  hundredths ;  74  and 
19  thousandths;  11  and  204  thousandths.  133.333 

8.  Add  45  and  19  thousandths;  7  and  71  hundred- 
thousandths  ;  93  and  4327  ten-thousandths ;  6  and  401 
ten-tiiousandths.  151.49251 

9.  Add  432  and  432  thousandths;  61  and  793  ten- 
thousandths;  100  and  7794  hundred-thousandths;  6.009; 
1000  and  1001  ten-thousandths.  1599.69834 

10.  Add  16  and  41  thousandths;  9  and  94  millionths ; 
33  and  27  hundredths;  8  and  969  thousandths;  32  and 
719906  milUonths.  100. 

11.  Add  204  and  9  ten-thousandths;  103  and  9  hun- 
dred-millionths;  42  and  9099  millionths;  430  and  99 
hundredths ;  220.0000009.  999.99999999 

12.  Add  35  ten-thousandths;  .00035;  35  millionths,  and 
35  ten-milHonths.  .0038885 

SUBTRACTION    OF    DECIMALS. 

144.     Subtraction  of  Decimals  is  the  process  of  find 
ing  the  difference  between  two  decimal  numbers. 

1.  From  729.835  subtract  461.5738. 

Solution. — Write  the  numbers  so  that  the  two 
decimal  points  may  be  in   a  column,  the  units  9       operation. 
and  1  in  the  first  column  to  the  left,  the  tenths  8       7  2  9.835 
and  5  in  the  first  column  to  the  right,  etc.;  then,       4  61.5  7  38 
subtracting  as  in  simple  numbers,  place  the  deci-        2  6  8.26  12 
mal    point    in    the    remainder    between    8    and    2 
under  the  column  of  decimal  points. 


180  RAY\S  NEW   PRACTICAL  ARITHMETIC. 

Rem. — The  ten-thousandth  phice  in  the  minuend  may  be  regarded 
as  occupied  by  a  cipher  (Art.  140). 

Rule. — 1.  Write  the  numbers  so  that  the  decimal  points 
and  figures  of  the  same  order  may  stand  in  the  same 
column. 

2.  Subtract  as  in  simple  numbers. 

3.  Place  the  decimal  point  in  the  remainder  under  the 
column  of  decimal  points. 

2.  From  97.5168  subtract  38.25942.  59.25738 

3.  From  20.014  subtract  7.0021.  13.0119 

4.  From  5.03  subtract  2.115.  2.915 

5.  From  24.0042  subtract  13.7013.  10.3029 

6.  From  170.0035  subtract  68.00181.  102.00169 

7.  From  .0142  subtract  .005.  .0092 

8.  From  .05  subtract  .0024.  .0476 

9.  From  13.5  subtract  8.037.  5.463 

10.  From  3  subtract  .00003.  2.99997 

11.  From  29.0029  subtract  19.003.  9.9999 

12.  From  5  subtract  .125.  4.875 

13.  From  1  thousand  subtract  1  ten-thousandth. 

999.9999 

14.  From  1  subtract  1  millionth.  .999999 

15.  From  25  thousandths  take  25  millionths.         .024975 


MULTIPLICATION  OF  DECIMALS. 

145.  Multiplication   of  Decimals   is   the   process    of 
finding  the  product  of  numbers  involving  decimals. 

146.  Placing    the    decimal    point   in    the   product  de- 
pends upon  the  following 


DECIMAL  FRACTIONS. 


181 


Principle. 

The  nvmber  of  decimal  orders  in  the  product   is  equal  to 
the  number  of  decimal   orders   in  both  the  factors. 


Thus,  let  the  factors  be  .2  and  .03;  then,  the  number  of  decimal 
orders  in  the  product  will  be  three.  For,  .2=z^q  and  .0o  =  y3_.  then, 
the  product  of  .2  by  .03  will  be  the  same  as  the  product  of  ^^  by  y^^ 


But, 


rX 


Too  —  To^o> 


and 


T^oo  = 


:.006.     Therefore,  .2X-03- 


.006, 


in  which  there  are  three  decimal  orders. 


Examples. 
147.     1.  Multiply  2.149  by  6.34. 

Solution. — Multiply  as  in  simple  numbers,  2140 
by  634. 

There  are  three  decimal  orders  in  2.140,  and  two 
decimal  orders  in  6.34;  hence,  there  must  be  five 
decimal  orders  in  the  product  (Art.  146).  There- 
fore, the  product  is  13.62466. 


OPERATION. 

2.1  4  0 
6.3  4 
"85  0  6 
6447 
1  2804 


]  3.6  2  4  6  6 


2.  Multiply  .0276  by  .035. 

Solution. — Multiply  the  numerator  (Art.  134,  6) 
276  by  the  numerator  35;  the  result  is  0660.  There 
are  four  decimal  orders  in  .0276,  and  three  decimal 
orders  in  .035;  hence,  there  must  be  seven  decimal 
orders  in  the  product  (Art.  146);  three  ciphers,  then, 
must  be  prefixed  to  0660.  Therefore,  omitting  the  .0000660 
cipher  on  the  right  (Art.  140, 2)  the  product  is  .000066. 


3.  Multiply  2.075  by  100. 

Solution. — Write    2075    and    place    the    decimal     operation. 
point  between  7  and  5,  two  places  farther  to  the  right  2  0  7.5 

than  it  is  in  2.075. 


182 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Rem. — To  multiply  207.5  by  100,  annex  a  cipher  and  mcve  the 
decimal  point  two  places  to  the  right. 

Kule, — 1.  Multiply  together  the  numerators  of  the  deci- 
mals as  in  Simple  Numbers. 

2.  Point  off  as  many  decimal  orders  in  the  jiroduct  as 
there  are  decimal  orders  in  both  factors. 

Rem.  1. — When  the  number  of  figures  in  the  product  of  the 
numerators  is  less  than  the  number  of  decimal  orders  required,  prefix 
ciphers.     (Ex.  2.) 

Rem.  2. — After  placing  the  decimal  point,  omit  ciphers  at  the 
right  of  the  decimal  part  of  the  product.     (Ex.  2.) 

Rem.  3. — To  multiply  a  decimal  by  10,  100,  1000,  etc.,  remove 
the  decimal  point  as  many  places  to  the  right  as  there  are  ciphers  in 
the   multiplier.      If   there    be    not    enough    figures    annex  ciphers. 


4.  Multiply 

5.  Multiply 

6.  Multiply 

7.  Multiply 
Multiply 
Multiply 
Multiply 

11.  Multiply 

12.  Multiply 

13.  Multiply 

14.  Multiply 
Multiply 
Multiply 
Multiply 
Multiply 
Multiply 
Multiply 

21.  Multiply 

22.  Multiply 


8. 

0, 

10 


15 
IG, 
17, 
18. 
19. 
20. 


33.21  by  4.41. 
32.16  by  22.5. 
.125  by  9. 
.35  by  7. 
.2  by  .8. 
.02  by  .4. 
.15  by  .7. 
125.015  by  .001. 
.135  by  .005. 
1.035  by  17. 
19  by  .125. 
4.5  by  4. 
.625  by  64. 
61.76  by  .0071. 
1.325  by  .0716. 
4.87  by  10. 
5.3  by  100. 
17.62  by  100. 
1.01  bv  10. 


146.4561 

723.6 

1.125 

2.45 

.16 

.008 

.105 

.125015 

.000675 

17.595 

2.375 

18. 

40. 

.438496 

.09487 

48.7 

530. 

1762. 

10.1 


DECIMAL  FKACTIONS.  183 

23.  Multiply  .0001  by  100.  .01 

24.  Multiply  1  tenth  by  1  hundredth.  .001 

25.  Multiply  1  hundred  by  1  ten-thousandth.  .01 

26.  Multiply  43  thousandths  by  21  ten-thousandths. 

.000090a 

27.  Multiply  40000  by  1  millionth.  .04 

28.  Multiply  .09375  by  1.064.  .09975 

DIVISION   OF   DECIMALS. 

148.  Division  of  Decimals  is  the  process  of  finding 
the  quotient  of  two  numbers  involving  decimals. 

149.  Placing  the  decimal  point  in  the  quotient  de- 
pends upon  the  following 

Principle. 

The  number  of  decimal  orders  in  the  quotient  is  equal  to 
the  number  of  decimal  orders  in  the  dividend^  less  the  num- 
ber in  the  divisor. 

Thus,  let  .006  be  divided  by  .03;  then,  the  number  of  decimal 
orders  in  the  quotient  will  be  one.  For  .006=:y^%^  and  .0?)=:j|}o; 
then,  the  quotient  of  .006  by  .03  will  be  the  same  as  the  quotient  of 
To%o  divided  by  ^f^.  But, \o%o  -  if o  =  I'o ;  ^^^^^l  -^,=  :2.  There- 
fore, .006 -^- .03  =  .2,  in  which  there  is  one  decimal  order. 

Examples. 

150.  1.  Divide  2.125  by  .5. 

Solution. — Divide  as  in  simple  numbers  2125  by  5.       operation. 
There   are   three   decimal    orders   in    2.125,   and   one         .5)2.12  5 
decimal  order  in  .5;  hence,  there  must  be  two  decimal  4.2  5 

orders    in   the   quotient    (Art.  149).     Therefore,  the 
quotient  is  4.25. 


184  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  Divide  .048  by  .006. 

Solution. — Divide  the  numerator  (Art.  134,  6)  48  operation. 
by  the  numerator  6.  There  are  three  decimal  orders  .006).0  4  8 
in    .048,  and    three    decimal    orders   in    .006;   hence,  8 

there  will  be  no  decimal  orders  in  the  quotient  (Art. 
149).     Therefore  the  quotient  is  8. 

3.  Divide  .3  by  .004. 

OPERATION. 

Solution. — Annex  two  ciphers  to  .3;  then  solve  .004  ).800 
as  in  Ex.  2.  75" 

4.  Divide  83.1  by  4. 

Solution. — Annex    two    ciphers   to   the    decimal      operation. 
(Art.  140,  1)  in  order  that  the  division  may  be  per-      4  )  8  3.1  00 
formed  exactly;  then  solve  as  in  Ex.  1.  20.7  7  5 

5.  Divide  2.11  by  3. 

Solution. — Annex  one  or  more  ciphers  to  the  dec-      operation. 
imal  (Art.  140,  1)  in  order  to   carry  the  division  as         3)2.1  10 
far  as  is  wanted;  then  solve  as  in  Ex.  1.  ,7  03-|- 

6.  Divide  475.(1  by  100. 

Solution.  —  Write  4756  and  place  the  decimal  operation. 
point  between  4  and  7,  two  places  farther  to  the  left  4.7  5  6 

than  it  is  in  475.6. 

Rem. — To  divide  4.756  by  100  prefix  a  cipher;  thus,  .04756. 

Rule. — 1.  Divide  the  numerator  of  the  dividend  by  the 
numerator  of  the  divisor  as  in  simple  numbers. 

2.  Point  off  as  many  decimal  orders  in  the  quotient  as 
the  number  of  orders  in  the  dividend  exceeds  the  number 
in    the    divisor. 


DECIMAL  FKACTIONS. 


185 


Eem.  1. — When  the  number  of  decimal  orders  in  the  dividend  is 
the  same  as  the  number  in  the  divisor,  the  quotient  is  an  integer 
(Ex.  2). 

Rem.  2. — When  the  number  of  decimal  orders  in  the  dividend  is 
less  than  the  number  in  the  divisor,  for  convenience  in  pointing  off, 
make  them  the  same  by  annexing  ciphers  to  the  dividend  (  Ex.  3). 

Rem.  8. — When  the  division  is  not  exact,  it  may  be  continued  to 
any  required  number  of  decimal  places  (Ex.  5). 

Rem.  4. — To  divide  a  decimal  by  10,  100,  1000,  etc.,  remove  the 
decimal  point  as  many  places  to  the  left  as  there  are  ciphers  in  the 
divisor.     If  there  be  not  enough  figures,  prefix  ciphers  (Ex.  6,  Rem). 


7. 

Divide  1.125  by  .03. 

37.5 

8. 

Divide  86.075  by  27.5. 

3.13 

9. 

Divide  24.73704  by  3.44. 

7.191 

10. 

Divide  206.166492  by  4.123. 

50.004 

11. 

Divide  .96  by  .24. 

4. 

12. 

Divide  .0425  by  .0085. 

5. 

13. 

Divide  21  by  .5. 

42. 

14. 

Divide  2  by  .008. 

250. 

15. 

Divide  37.2  by  5. 

7.44 

16. 

Divide  100.8788  by  454. 

.2222 

17. 

Divide  .000343  by  3.43. 

.0001 

18. 

Divide  9811.0047  by  .108649. 

90300. 

19. 

Divide  .21318  by  .19. 

1.122 

20. 

Divide  102048  by  .3189. 

320000. 

21. 

Divide  .102048  by  3189. 

.000032 

22. 

Divide  9.9  by  .0225. 

440. 

23. 

Divide  872.6  by  100. 

8.726 

24. 

Divide  4.5  by  1000. 

.0045 

25. 

Divide  400  by  10000. 

.04 

26. 

Divide  1  tenth  by  10. 

.01 

27. 

Divide  1  by  1  tenth. 

10. 

28. 

Divide  10  by  1  hundredth. 

1000. 

29. 

Divide  1.7  by  64. 

.0265625 

186  liAY'S  NEW  PRACTICAL  ARITHMETIC. 

30.  Divide  .08  by  80.  .001 

31.  Divide  1.5  by  7.  .2142857  + 

32.  Divide  11.1  by  32.76.  .3388278  + 

33.  Divide  .0123  by  3.21.  .00383177  + 

DECIMAL  COMPOUND  NUMBERS. 

151.  A    decimal    is    reduced  to  a  lower  denomination 
by  multiplication  (Art.  63,  Rule  I). 

1.  Reduce  .05  gal.  to  the  decimal  of  a  pint. 

orKRATION. 

.06 

Solution. — To  reduce  .05  gal.  to  the  decimal  of  a  4 

pint,  multiply  by  4  and  by  2.     The  result  is  .4  pint.  .20 

2_ 
.4 

2.  Reduce  .035  pk.  to  the  decimal  of  a  pint.        .56  pt. 

3.  Reduce  .0075  bu.  to  the  decimal  of  a  quart.      .24  qt. 

4.  Reduce  .005  yd.  to  the  decimal  of  an  inch.       .18  in. 

5.  Reduce  .00546875    A.    to    the    decimal    of  a    square 
rod.  .875  sq.  rd. 

152.  To  find  the  value  of  a  decimal  in  integers  (Art. 
125). 

1.  Find  the  value  of  .3125  bu.  in  integers.  ^ 

OPERATION. 

Solution. — To  reduce  .3125  bu.  to  pecks,  multiply  .8  12  5 

by  4;    the  result  is  1.25  pk.     To  reduce  .25   pk.  to  4 

quarts,  multiply  by  8;  the  result  is  2  qt.     Therefore,  1.2  5  00 

.3125  bu.  equals  1  pk.  2  qt.  8 

2:00 

2.  Find  the  value  of  .75  yd.  in  integers.  2  fl.  3  in. 

3.  Find  the  value  of  .3375  A.  in  integers.        54  sq.  rd. 


DECIMAL  FRACTIONS.  187 

4.  Find  the  value  of  .7  lb.  Troy  in  integers. 

8  oz.  8  pwt. 

5.  Find  the  value  of  .8125  bii.  in  integers. 

3  pk.  2  qt. 

6.  Find  the  value  of  .44  mi.  in  integers. 

140  rd.  4  yd.  1  ft.  2.4  in. 

7.  Find  the  value  of  .33625  cwt.  in  integers. 

33  lb.  10  oz. 

153.  A  decimal    is  reduced   to  a  higher  denomination 
by  division  (Art.  63.     Rule  II). 

1.  Reduce  .64  pt.  to  the  decimal  of  a  gallon. 

OPERATION. 

Solution. — To  reduce  .64  pt.  to  the  decimal  of  a  2).6  4 

gallon,  divide  by  2  and  by  4.     The  result  is  .08  gal.  4  ).3  2 

708 

2.  Reduce  .72  qt.  to  the  decimal  of  a  bushel. 

.0225  bu. 

3.  Reduce  .77  yd.  to  the  decimal  of  a  mile. 

.0004375  mi. 

4.  Reduce  .25  pt.  to  the  decimal  of  a  gallon. 

.03125  gal. 

5.  Reduce  .6  pt.  to  the  decimal  of  a  bushel. 

.009375  bu. 

6.  Reduce  .7  rd.  to  the  decimal  of  a  mile. 

.0021875  mi. 

Promiscuous  Examples. 

154.  1.  What    is    the    cost   of  9  yd.  flannel,  at    $0.40 
per  yard,  and  12  yd.,  at  S0.75  per  yard?  S12.60. 

2.  What  is  the  cost  of  2.3  yd.  of  ribbon,  at  $0.45  per 
yard,  and  1.5  yd.,  at  $0,375  per  yard?  $1.5975. 


188  KAYS  NEW  PKACTICAL  ARITHMETIC. 

3.  What  is  the  cost  of  16.25    yd.  of  cloth,  tit  $2.6875 
per  yard?  $43.671875. 

4.  At  $0.75  per  bushel,  how  much  wheat  can  be  boui^ht 
for  $35.25?  47  bii. 

5.  At  $2.5625  per  yard,  how  much  cloth  can  be  bought 
for  $98.40  ?  38.4  yd. 

6.  What  will  6  cwt.  50  lb.  of  hops    cost    at   $3.25  per 
hundred-weight?  $21,125. 

7.  What    will    14    bu.  3    pk.    4    qt.    of    corn    cost,   at 
$0,625  per  bushel?  $9.296875. 

8.  What  will  13  A.  115  sq.  rd.  of  land  cost,  at  $17.28 
per  acre?  $237.06. 

9.  At    $0.3125    per    bushel,    how    much    corn    can    be 
bought  for  $9.296875?  29  bu.  3  pk. 

10.  At  $4.32  per  acre,  how  much  land    can  be  bought 
for  $59,265?  13  A.  115  sq.  rd. 

11.  If  63  gal.  of  wine  cost  $49,  what  will  464  gal  cost? 

$360.88  + 

12.  Add  .34  3^d.,  1.07  ft.  and  8.92  in.  2  ft.  10  in. 

13.  Add  .625  gal.  and  .75  qt.  3  qt.  .5  pt. 

14.  From  1.53  yd.  subtract  2  ft.  3.08  in.         2  ft.  4  in. 

15.  From  .05  yr.  subtract  .5  hr. 

18  da.    5   hr.  48  min. 

16.  From  .41  da.  subtract  .16  hr. 

9  hr.  40  min.  48  sec. 

17.  Find  the  value  of  .3  yr.  in  integers. 

109  da.  13  hr.  48  min. 

18.  What    is   the  cost  of  343  3^d.  2  ft.  3  in.  of  tubing, 
at  $0.16  per  yard?  $55. 

19.  At    $690.35    per    mile,  what    is  the  cost  of  a  road 
17  mi.  135  rd.  long?  $12027.19140625. 


THEjMETRreiSYSTEM. 


DEFINITIONS. 

155.  1.  The  Metric  System  is  so  called  from  the 
meter,  the  unit  upon  which  the  system  is  based. 

Eem. — The  French  originated  this  system  of  weights  and  measures 
at  the  close  of  the  last  centurj^,  and  its  use  in  France  became 
obligatory  in  1841.  The  metric  system  is  now  legal  in  nearly  all 
civilized  countries,  and,  in  several,  it  is  making  its  way  rapidly  into 
general  use.  In  1866,  its  use  was  legalized,  in  the  United  States,  by 
act  of  Congress.  It  is  in  general  use  by  scientific  men  throughout 
the  world. 

2.  All  the  units  of  the  other  measures  are  derived  in 
a  simple  manner  from  the  meter.     Thus, 

1st.  The  Meter  is  the  unit  of  Length.  It  is  the  base 
of  the  Metric  System,  and  is  very  nearly  one  ten-mill- 
ionth (.0000001)  part  of  the  quadrant  extending  through 
Paris  from  the  equator  to  the  pole. 

2d.  The  Ar  is  the  unit  of  Land  Measure.  It  is  a 
square  whose  side  is  10  meters. 

3d.  The  Liter  is  the  unit  of  capacity.  It  is  a  vessel 
wdiose  contents  are  equivalent  to  a  cube  the  edge  of 
which  is  .1  meter. 

4th.  The  Gram  is  the  unit  of  Weight.  It  is  the 
weight  of  a  cube  of  pure  water  whose  edge  is  .01   meter. 

(189) 


190 


KAY'S  NEW  rilACTlCAL  ARITHMETIC. 


3.  The  name  of  each    denomination    indicates  at   once 
its  relation  to  the  unit  oi'  the  measure. 


27 


Thus:  1st.  The  luinies  uf  the  louver  dononiiiiutions 
are  formed  by  prefixing  to  the  natiie  of  the  unit  the 
Latin  numerals  mdli  (.001),  ctntl  (.01),  and  dcci  (.1). 

For  example,  a  miUimeter  is  one  thousandth  of  a 
meter;  a  centigi-a^n  is  one  hundredth  of  a  gram;  and 
a  deciliter  is  one  tenth  of  a  liter. 

2d.  The  names  of  the  liiffhrr  drnoiiiinatidiis  arc 
formed  by  prefix!  Hi;-  to  tin-  unii  tin-  a  ml;  miiiuTal.-  (Iil;ii 
(10),  heldo  (100),  /.v/o  i  l(»n(»  .  mimI  inur'in  i  KMHXh.  For 
example,  a  dekameter  i-  I'^n  nirt.T.-;  a  hrl;ii_,r,frr  i>  ..nc 
hundred  liters;  a  kilocjrain  is  one  tliousand  grams;  and 
a  >ni/ria7neter  is  ten  thousand  meters. 

4.  Since  in  the  Metric  System  10,  100.  1000, 
etc.,  units  of  a  lower  denomination  make  a 
unit  of  a  higher  denomination,  it  follows  thai. 

Ist.  A  number  is  reduced  in  <i  i.owf.h  d,  nom- 
ination by  removing  the  drci/jud  j/oi/it  d.s  innny 
places  to  the  right  as  there  are  ciphers  in  the 
multiplier. 

2d.  A  number  is  reduced  to  a  iiKiriEu  drncnn- 
ination  by  removing  the  deciniai  jxiinf  a.^  iiKinij 
places  to  the  left  as  there  are  ciphers  in  the 
divisor. 


MEASURES  OF  LENGTH. 

156.     The   Meter    is  the  unit  of  length  ;  it  is  legal  in 
the  United  States  at  39.37  inches. 


Rem.  1.— Its  length  is  also  a  little  less  than  1.1  yards,  or  nearly 
3  ft.  3|  in.,  which  may  be  remembered  as  the  rule  of  ike  three  threes. 


decimeter, 

"       dm. 

meter. 

"        m. 

dekameter, 

^'       Dm. 

hektometer 

"       Hm. 

kilometer, 

"       Km. 

myriameter. 

"       Mm, 

THE  METKIC  SYSTEM.  If 

Rem.  2. — The  decimeter  and  its  divisions  are  shown  in  the  en- 
graving on  the  opposite  page. 

Rem.  3. — Standard  meters  have  l)een  provided  by  the  United 
States,  and  copies  have  been  furnished  to  the  several  states. 

Table. 

10  millimeters,  marked  mm.,  are  1  centimeter,  marked  cm. 

}0  centimeters 

JO  decimeters 

10  meters 

10  dekameters 

10  hektometers 

10  kilometers 

Rem. — The  measures  chiefly  used  are  the  meter  and  kilometer. 
The  meter,  like  the  yard,  is  used  in  measuring  cloth  and  short  dis- 
tances; the  kilometer  is  used  in  measuring  long  distances. 

1.  Eeduce  5.638  m.  to  centimeters. 

Solution. — To  reduce  meters  to  centimeters,  multiply  by  100. 
"Write  5638  and  place  the  decimal  point  between  3  and  8,  two 
orders  farther  to  the  right  than  it  is  in  5.638  (Art.  155,  4,  1st). 

A71S,  563.8  cm. 

2.  Eeduee  3642.9  m.  to  kilometers. 

Solution.  —  To  reduce  meters  to  kilometers,  divide  by  1000. 
"Write  36429  and  place  the  decimal  point  between  3  and  6,  three 
orders  farther  to  the  left  than  it  is  in  3642.9  (Art.  155,  4,  2d). 

Ans.  3.6429  Km. 

3.  Reduce  4.27  Dm.  to  centimeters. 

Solution. — To  reduce  dekameters  to  centimeters,  multiply  by 
10  X  100  =  1000.     "Write  427  and  annex  a  cipher  (  Ex.  1  ). 

A71S.  4270  cm. 

4.  Reduce  5.6  dm.  to  hektometers. 

Solution. — To  reduce  decimeters  to  hektometers,  divide  by  10  X 
100  =  1000.  Write  56,  prefix  two  ciphers,  and  place  the  decimal 
point  before  them  (Ex.  2).  A71S.  .0056  Hm. 


J92  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

5.  Eeduce  30.75  m.  to  centimeters.  3075  cm. 

6.  Eeduce  4.5  Km.  to  meters.  4500  m. 

7.  Reduce  75  mm.  to  meters.  .075  m. 

8.  Reduce  .025  Dm.  to  decimeters,  2.5  dm. 

9.  Reduce  36.5  dm.  to  dekameters.  .365  Dm. 
10.  Reduce  .4875  Km.  to  centimeters.  48750  cm. 

LAND   OR    SQUARE    MEASURE. 

157.  The  Ar  is  the  unit  of  Lund  Measure;  it  is  legal 
at  119.6  square  yards. 

Table. 

100  centars,  marked  ca.,  are   1  ar,     marked  a. 
100  ars  "      1  hcktar,    "        Ha. 

Rem.  1. — An  ar  is  100  square  meters,  marked  m^,  Xhe  hektar 
is  very  nearly  2|  acres. 

Rkm.  2.— For  measuring  other  surfaces,  squares  of  the  meter  and 
its  subdivisions  are  used. 

1.  Reduce  2.625  a.  to  centars.  262.5  ca. 

2.  Reduce  397.8  a.  to  hektars.  3.978  Ha. 

3.  Reduce  2500  ca.  to  hektars.  .25  ITji. 

4.  Reduce  3.8  a.  to  square  meters.  380  m^. 

MEASURES    OF    CAPACITY. 

158.  The  Liter  is  the  unit  of  Capacity:  it  is  legal  at 
1.0567  quarts,  Liquid  measure. 

Table. 

10  centiliters,  marked  cl.,  are  1  deciliter,  marked  dl. 

10  deciliters  ♦*  1  liter,  "       1. 

10  liters  "  1  dekaliter,       ••        Dl. 

10  dekaliters  "  1  hektoliter,    "        HI. 


THE  METRIC  SYSTEM.  103 

Rem.  1.— The  measures  commonly  used  are  the  liter  and  hektoliter. 
The  liter  is  very  nearly  a  quart;  it  is  used  in  measuring  milk,  wine, 
etc.,  in  moderate  quantities.  The  hektoliter  is  about  2  bu.  3i  pk.;  it 
is  used  in  measuring  grain,  fruit,  roots,  etc.,  in  large  quantities. 

Rem.  2. — Instead  of  the  milliliter  and  the  kiloliter,  it  is  customary 
to  use  the  cubic  centimeter  and  the  cubic  meter  (marked  m^),  which 
are  their  equivalents. 

Rem.  3. — For  measuring  wood  the  ster  is  used;  it  is  a  cubic  meter. 

1.  Eeduce  2.456  1.  to  centiliters.  245.6  cl. 

2.  Reduce  873.5  1.  to  hektolitem  8.735  HI. 

3.  Eeduce  1.83  HI.  to  deciliters.  1830  dl. 

4.  Eeduce  2400  cl.  to  dekaliters.  2.4  Dl. 

5.  Eeduce  1400  1.  to  cubic  meters.  1.4  m^ 

MEASURES   OF  WEIGHT. 

159.  The  Gram  is  the  unit  of  Weight;  it  is  legal  at 
15.432  grains. 

Table. 


10  milligrams,  marked  mg.,  are 

1  centigram,  marked 

[  eg. 

10  centigrams                              " 

1  decigram,          " 

<jg- 

10  decigrams                                " 

1  gram,            " 

g- 

10  grams                                         " 

1  dekagram,        " 

i>g- 

10  dekagrams                               " 

1  hektogram,      " 

Hg. 

10  hektograms                              *' 

1  kilogram,          " 

Kg. 

10  kilograms                                 " 

1  myriagram,      " 

Mg. 

10  myriagrams                             " 

1  quintal,              " 

Q 

10  quintals,  or  1000  kilograms,  " 

1  metric  ton,        " 

M.T. 

Rem. — The  weights  commonly  used  are  the  gram,  kilogram,  and 
metric  ton.  The  gram  is  used  in  mixing  medicines,  in  weighing  the 
precious  metals,  and  in  all  cases  where  great  exactness  is  required. 
The  kilogram — or,  as  it  is  commonly  called,  the  "kilo"— is  the  usual 
weight  for  groceries  and  coarse  articles  generally;  it  is  very  nearly 
2i  pounds  Av.  The  metric  ton  is  used  for  weighing  hay  and  other 
heavy  articles;  it  is  about  204  lb.  more  than  our  ton. 
Prac.  13. 


194 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


1.  Eediice  1428.06  g.  to  kilograms.  1.42806  Kg. 

2.  Eeduco  .28  Kg.  to  grams.  280  g. 

3.  Reduce  1713.5  Kg.  to  metric  tons.  1.7135  M.T. 

4.  lieduce  .00654  Hg.  to  centigrams.  65.4  eg. 

5.  Reduce  192.7  dg.  to  dekagrams.  1.927  Dg, 

IGO.  The  legal  and  approximate  values  of  those  de- 
nominations of  the  Metric  System  which  are  in  common 
use  are  presented  in  the  following 

Table : 


DENOMINATION. 

LEGAL   VALUE. 

APP.    VALUE. 

Meter. 

39.37  inches. 

3  ft.  3|  inches. 

Kilometer. 

.62137  mile. 

f  mile. 

Square  Meter. 

1.196  sq.  yards. 

10|  sq.  feet. 

Ar. 

119.6  sq.  yards. 

4  sq.  rods. 

Hectar. 

2.471  acres. 

21  acres. 

Cubic  Meter. 

1.308  cu.  yards. 

35J  cu.  feet. 

Ster. 

.2759  cord. 

\  cord. 

Liter. 

1.0567  quarts. 

1  quart. 

Hektoliter. 

2.8375  bushels. 

2  bu.  3J  pecks. 

Gram. 

15.432  gr.  T. 

15^  grains. 

Kilogram. 

2.2046  lb.  Av. 

2^  pounds. 

Tonneau. 

2204.6  lb.  Av. 

1  T.  204  lb. 

Note. — The  legal  value  is  used  in  solving  the  following  examples. 
1.  How  many  yards,  feet,  etc.,  in  4  m.  ? 


Solution. — In  4  meters  there  are  4  times 
39.37  in.  which  are  157.48  in.,  157.48  in. 
reduced  to  integers  of  higher  denomina- 
tions are  4  yd.  1  ft.  1.48  in. 


OPERATION. 

39.37 

4 

12)157.48 

3)13  ft.  1.4  8  in 

4  yd.      1   ft. 

THE  METRIC  SYSTEM.  I95 

2.  What  is  the  value  of  36  lb.  in  kilograms? 

OPERATION. 

Solution.— In  36  pounds  there  2.2  0  4  G  )  3  6.0  0  0  0  (  1  G.3  2  9  + 
are   as  many  kilograms  as  2.2046  2  2046 

are   contained  times  in  36  which  139540 

are   16.329  +  .  1322  7  6 


72640 
66138 


65020 
44092 
2  0  9  2  8  0 
198414 

3.  What  is  the  value  of  20  Km.  ?  12.4274  mi. 

4.  How  many  hektars  in  160  acres?  64.75-|-  Ha. 

5.  What  is  the  value  of  49  m.?         9  rd.  4  yd.  3.13  in. 

6.  What  is  the  value  of  15  g.?  9  pwt.  15.48  gr. 

7.  How  many  hektoliters  in  42  bu.?  14.8+  HI. 

8.  How  many  cords  in  500  sters?  137.95  C. 

9.  How  many  square  yards  in  a  roll  of  paper  9  m. 
long  and  .5  m.  wide?  5.382  sq.  yd. 

10.  32  1.  are  how  many  gallons?  8.4536  gal. 

Miscellaneous  Examples. 

161.  1.  What  is  the  sum  of  127  dl.,  4.87  1.,  1563  cl, 
and  234.5  dl.?  56.65  1. 

2.  What  will  be  the  cost  of  45  Ha.  of  land,  at  $3.32 
an  ar?  S14940. 

3.  A  merchant  paid  $457.92  for  cloth,  at  S3  a  meter: 
how  many  meters  did  he  buy?  152.64  m. 

4.  A  block  of  marble  .72  m.  long,  .48  m.  wide,  and  .5 
m.  thick  cost  $.864 :  what  is  the  cost  of  the  marble  per 
cubic  meter?  $5.' 


196  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

5.  A  manufacturer  bought  380  stcrs  of  wood  for 
$454.10:    how  much  was  that  a  8tcr?  SI. 195 

6.  How  many  hektoliters  of  oats  in  4685  sacks,  each 
containing  1.6  HI.?  7496  HI. 

7.  I  bought  346.75  Kg.  of  coffee  for  $194.18  :  what 
did  I  pay  per  kilogram?  $0.56 

8.  The  nickel  5-cent  coin  weighs  5  g.  and  is  2  cm. 
in  diameter:  what  would  be  the  weight  of  enough  of 
these  coins  hiid  in  a  row,  to  make  a  meter  in  length  ? 

250  g. 

9.  How  much  lining  1.85  m.  wide  will  it  take  for  a 
garment  made  of  6.5   m.  of  cloth   1.25  m.  wide? 

4.39+  m. 

10.  How  many  kilometers  from  Cincinnati  to  Dayton, 
the  distance  being  60  miles.  96.56-|-  Km. 

11.  A  map  is  29  mm.  long  and  22.4  mm.  wide :  what 
space  does  it  cover?  649.6  mm^. 

12.  The  distance  between  two. towns  is  13.24037  Km.: 
how  many  steps  of  .715  m.  each,  must  I  take  to  walk 
that  distance  ?  18518  steps. 

Note. — To  illustrate  the  difference  between  the  metric  system  and 
our  common  system  of  measures,  a  similar  example  may  be  given, 
substituting  8  mi.  72  rd.  4  yd.  1.7  in.  for  the  distance,  and  28.15  in. 
for  the  length  of  one  step. 


162.  1.  Any  per  cent  of  a  number  is  so  msiuy  hun- 
dredths of  it. 

Thus,  1  per  cent  of  a  number  is  yi^  of  it,  2  per  cent  is  y^^,  etc. 

Rem. — Per  cent  i^  from  the  Latin  per  centum,  by  the  hundred. 

2.  The  sign  of  per  cent  is  %,  read  per  cent. 
Thus,  5  %  is  read  five  per  cent. 

3.  In  all  operations  with  per  cent,  it  may  be  expressed 
in  two  ways:  1st.  As  a  common  fraction;  2d.  As  a 
decimal. 

Thus  the  following  expressions  are  equivalent: 


One 

per  cent. 

1/., 

is 

T^t> 

or 

.01 

Two 

per  cent, 

2%, 

is 

TOO  ~  A' 

or 

.02 

Three 

per  cent. 

3f., 

is 

Too 

or 

.03 

Four 

per  cent. 

4/„ 

is 

TOO  =  TJ' 

or 

.04 

Five 

per  cent, 

5^„ 

is 

Too  —    2^' 

or 

.05 

Six 

per  cent, 

6fc, 

is 

t!  0  =  A' 

or 

.06 

Rem.  1. — Per  cent,  which  is  expressed  as  a  mixed  number,  may  be 

reduced   to   equivalent   expressions   by  Arts.  121   and  142.     Thus, 

4i  9 

4^  ^  =  Tqq»  which  may  be  reduced  to  ^qqI    also,  i\  ^  =  .045. 

no7) 


198 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


Express    the    following    as    common    fractions    and    as 
decimals : 


1. 
2. 
8. 
4. 
5. 
6. 


10% 

20% 
30<^ 


70 
2i% 


yV  and  .10 

^-^   and  .15 

i  and  .20 

f^  and  .30 

-1-  and  .50 
i^   and 


.025 


7. 

^>i% 

tV  a^''^ 

.0625 

8. 

12i% 

-J-    and 

.125 

9. 

18|% 

A  and 

.1875 

10. 

37i% 

f   and 

.375 

11. 

56i% 

A  and 

.5625 

12. 

874% 

1   and 

.875 

Rem.  2.--C()inmon   fractions    may  be   reduced  to  hundredths  by 
Art.  108,  snul  tluMi  read  as  per  cent.     Thus,  ^  =  .165  or  16§  ol 

How    many  per    cent   are  equivalent  to  the  following 
fractions  ? 


1. 

h- 

8% 

6.     \. 

25% 

2. 

A- 

12% 

7-     f 

40% 

3. 

^V 

16% 

8.     f. 

75% 

4. 

A-  ■ 

3i% 

9.     ^. 

33^% 

5. 

tV- 

8J% 

DEFINI 

10.     Jj. 

TIONS. 

43|% 

163.     1.   Percentage   embraces   the  various  operations 
with  per  cent. 

2.  In  Percentage    three  quantities  are  considered.     (1) 
the  Base^  (2)  the  Rate^  and  (3)  the  Percentage. 

3.  The  Base  is    the  number  upon  which    the  per  cent 
is  estimated. 

4.  The    Rate    is    the    per    cent    when    expressed    as    a 
common  fraction  or  as  a  decimal. 

5.  The    Percentage    is    the    resujt    of  taking   the   per 
cent  of  the  base. 

6.  Any  two  of  these  quantities  being  given,  the  third 
may  be  found.     There  are  four  cases. 


PEECENTAGE. 


199 


CASE    I. 

164.  Given  the  base  and  the  rate,  to  find  the  per- 
centage. 

1.  What  is  25%  of  32? 

OPERATION. 

SoLUTiox.— 25^0  is  i  (Art.  162).     i  of  32  is  8.  2b<f^  =  ^ 

2.  What  is  7%   of  162? 

OPERATION. 

Solution.— 7  ^^  is  .07  (Art.  162).     Multiplying  1G2  by         16  2 
.07,  the  result  is  11.34.  .0  7 

1  1.3  4 

Rule. — Multiply  the  base  by  the  rate;  the  product  will  be 
the  percentage. 

Rem. — Whether  the  rate  should  be  expressed  as  a  common  frac- 
tion, or  as  a  decimal,  must  be  a  matter  of  judgment.  That  form 
of  expression  is  best  which  is  simplest  or  most  convenient  in  the 
given  example. 


3.  What  is 

4.  What  is 

5.  What  is 

6.  What  is 

7.  What  is 

8.  What  is 

9.  What  is 

10.  What  is 

11.  What  is 

12.  Wliat  is 

13.  What  is 

14.  What  is 

15.  What  is 

16.  What  is 


1% 
2% 
3% 

^% 

3f% 
4% 

5% 

H% 

^% 

H% 

8% 

8i% 

10% 

12i% 


of  278? 
of  180? 
of  97? 
of  165? 
of  240? 
of  140? 
of  118? 
of  150? 
of  250? 
of  450? 
of  11? 
of  384? 
of  57? 
of  292? 


2.78 
3.6 

2.91 
5.5 
9. 
5.6 
5.9 
8. 
15. 
30. 
,88 
32. 
5.7 

36.5 


200 


HAY'S  NKW  PRACTICAL  AUITUMETIC. 


17. 
18. 
19. 
20. 
21. 
22. 
28. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 


What 
What 
What 
What 
What 
What 
What 
What 
What 
What 
What 
What 
What 
What 
What 


s  15% 
s     17% 

« m% 

8  20% 
H  25% 
8  33J% 
s  45% 
8     50% 

B         i% 

«     8% 

»  A% 

s  125% 
8  208% 
8  450% 
8  1000% 


of  95? 
of  53.4? 
of  11.2? 
of  9.85  ? 
of  43? 
of  0.93? 
of  5.7  ? 
of  38.75? 
of  456  ? 
of  464? 
of  144? 
of  36? 
of  650? 
of  12? 
of  24.75? 


14.25 

9.078 

2.1 

1.97 

10.75 

2.31 

2.565 

19.375 

2.28 

1.74 

.63 

45. 

1352. 

54. 

247.5 


CASE  II. 

165.     Given  the    base  and  the  percentage,  to  find  the 
rate. 


1.  What  per  cent  of  8  is  2? 
Solution.— 2  is  \  of  8  (Art.  120).     \  is  25^^. 

2.  What  per  cent  of  56  is  3.5? 


OPERATION. 


OPERATION. 

Solution. — Dividing  3.5  by  56,  the  result         3.5  -^  5  6  —..0  6  2  5 
IS  .0625.     .0625  is  6]  %.  .0  6  2  5  =r  6]  ^^ 

Explanation. — One  per  cent  of  56  is  .56;  then  3.5  is  as  many  per 
cent  as  .56  is  contained  times  in  3.5. 

-R^lQ, — 1.  Divide  the  percentage  by  the  base;  the  quotient 
will  be  the  rate. 


PERCENTAGE. 

201 

cent  of 

15 

is  3? 

20. 

cent  of 

50 

is  6? 

12. 

cent  of 

75 

is  4.5? 

6. 

cent  of 

9 

18  3? 

33J. 

cent  of 

25 

is  .25? 

1. 

cent  of  142.6 

is  7.13? 

5. 

cent  of 

9 

is  9? 

100. 

cent  of 

9 

is  13.5? 

150. 

cent  of 

243 

is  8.505? 

^. 

cent  of 

2 

s  .002? 

tV- 

cent  of 

3532 

is  13.245? 

I 

cent  of 

1 

i«  A? 

15. 

cent  of 

t 

•s  A? 

20. 

cent  of 

M 

S    -2_? 

37f 

cent  of 

llf 

IS  5^? 

45. 

cent  of 

57-1 

IS  lOf? 

18f. 

CASE   HI. 

3.  What  per 

4.  Wlaat  per 

5.  What  per 

6.  What  per 

7.  What  per 

8.  What  per 

9.  What  per 

10.  What  per 

11.  What  per 

12.  What  per 

13.  What  per 

14.  What  per 

15.  What  per 

16.  What  per 

17.  What  per 

18.  What  per 


166.     Given  the    rate    and    the  percentage  to  find  the 
base. 


1.     15  is  25%  of  what  number? 

Solution.— 25  ^,  is  |  (Art.  162).     Since  15  is  | 
of  some  number,  the  number  is  4  X  15  =  00, 


OPERATION. 

25^,  =  i 
15X4  =  60 


2.     4.93  is  17%   of  what  number? 


OPERATION. 

17/,  =.17 


Solution. — 17  oi  is  .17  (Art.  162).    Since  some 
number  multiplied  ])y  .17  gives  the  product  4.93,     4.9  3  -t-.I  7  =  29 
the  number  is  4.93  divided  bv  .17  or  29. 


Rule. — Divide   the  percentage   by  the   rate;    the   quotient 
will  be  the  base. 


202  RAY'S  NEW  PRACTICAL  ARITHMETIC. 


3. 

60 

is 

20% 

of  what  number? 

300. 

4. 

90 

is 

75% 

of  what  number? 

120. 

5. 

85 

is 

125% 

of  what  number? 

()8. 

6. 

7.13 

is 

23% 

of  what  number? 

31. 

7. 

20.23 

is 

34% 

of  what  number? 

59.5 

8. 

23.5 

is 

47% 

of  what  number? 

50. 

9. 

45 

is 

H% 

of  what  number? 

3000. 

10. 

2.25 

is 

12i% 

of  what  number? 

18. 

11. 

f 

is 

250% 

of  what  number? 

A- 

12. 

14^ 

is 

l«t% 

of  what  number? 

CASE   IV. 

85f. 

167.     Given  the   rate   and   the   sum   or  the   difference 
of  the  base  and  percentage,  to  find  the  base. 

1.  A  number,  plus  35%  of  itself,  equals  675:    what  is 
the  number? 

OPERATION. 

Solution. — 35  %  is  .35.     The  number  plus  35  ^  =.3  5 

.35  of  itself  equals  1.35  of  it;  then,  1.35  of  1  -f-.3  5=:1.3  5 

the  number  is  675,  and    the    number    itself  6  75-^-1.3  5  =  5  00 
is  G75  divided  by  1.35,  or  500. 

2.  A  number,  minus  5%  of  itself,  equals    57:   w^hat  is 
the  number? 

OPERATION. 

Solution.— 5  c!^  is  2V-     The  number  minus  ^^^  5  ^  —  ^i^ 

of  itself  equals  ^§  of  it;  then,  ig  of  the  number  \%  —  uV  =^  2^ 

is  57,  2^0  of  it  is  3,  and  the  number  is  20  times  3 

3  =  CO.  ,^  X  H  ^  60 

Rule. — Divide  the  sum  by  1   plus  the  rate,  or  divide  the 
difference  by  1  minus  the  rate;  the  quotient  will  be  the  base 

3.  721   is  3^  greater  than  a  certain  number;  what  is 
the  number?  700. 


PERCENTAGE.  203 

4.  68  is  66%  less  than  what  number?  200. 

5.  What  number,  increased  by  25%  of  itself,  amounts 
to  2125?  1700. 

6.  What  number,  diminished  by  6%  of  itself,  is  equal 
to  7.52?  8. 

7.  8250  is  37J%  greater  than  what  number?  6000. 

8.  What  fraction,  less  10%  of  itself,  equals  f  ?  -:f^. 

9.  6.6  is  20%   more  than  what  number?  5.5 

168.     Formulas  for  the  Four  Cases  of  Percentage. 

Let  b   represent    the   base,  r  the  rate,  and  j^  the   per- 
centage.    Then, 

Case      I.     b  X  r  =p. 
Case    II.    J)  ^b  =^r. 


Case  III.    ])  -^  r  ^=b. 

1  +  r  \—r 


Case  IV.     ^±£==6.     *-^^=  6. 


Miscellaneous  Examples. 

169.  1.  I  had  S800  in  bank  and  drew  out  36%  of  it: 
how  much  had  I  left?  $512. 

2.  A  man  had  $300;  after  he  had  spent  $225,  what 
per  cent  did  he  have  left?  25%. 

3.  A  merchant  withdrew  40%  of  his  deposits,  leaving 
$3000  remaining  in  the  bank  :  what  amount  did  he  with- 
draw? '  $2000. 

4.  A  grain  dealer  sold  corn  for  56  ct.  a  bushel,  which 
was  40%  more  than  it  cost  him:  what  was  the  cost  per 
bushel?  40  ct. 

5.  A  man  sold  a  horse  for  $175,  which  was  12J%  less 
than  the  horse  cost:  what  did  the  horse  cost?  $200. 


204  KAYS  NEW  PRACTICAL  ARITHMETIC. 

G.  A  grocer  bought  4  sacks  of  coffee  of  75  pounds  Ccach; 
12^^  was  lost  by  waste:  what  was  the  remainder  wortli 
at  35  cents  per  pound?  $91.87^. 

7.  A  man  owed  $500;  lie  paid  $425:  what  per  cent  of 
the  debt  remains  unpaid?  15%. 

8.  A  speculator  invested  75%  of  his  estate  in  bonds, 
and  the  remainder  of  it,  amounting  to  $5000,  in  real 
estate:  how  much  did  he  invest  in  bonds?  $15000. 

9.  A  farmer  owned  a  farm  of  250  A.  86  sq.  rd.,  which 
was  12^%  more  than  his  neighbor  owned:  how  much 
land  did  his  neighbor  own?  222  A.  112  sq.  rd. 

10.  A  flock  of  IGO  sheep  increased  35%  in  one  year: 
how  many  were  then  in  the  flock?  216. 

11.  A  miller  takes  for  toll  6  qt.  from  every  5  bu.  of 
wheat  ground:  what  per  cent  does  he  take?  3i%- 

12.  A  farmer  ow^ning  45%  of  a  tract  of  land,  sold  540 
acres,  which  was  G0%  of  what  he  owned:  how  many 
acres  were  there  in  the  tract?  2000  A. 

13.  When  the  gold  dollar  is  worth  7%  more  than  the 
greenback  dollar,  how^  much  in  gold  are  $371.29  in 
greenbacks  w^orth?  $347. 

'4.  A's  salary  is  $800  a  year;  he  spends  18%  of  it  for 
rent,  15%  for  clothing,  23%  for  provisions,  and  12%  for 
sundries:  how  much  does  he  save  annually?  $25G. 

15.  A  pupil  at  an  examination  answered  17  of  the  20 
questions  correctly :  what  per  cent  did  he  make  ?     85^ . 

16.  2  bu.  3  pk.  are  33J%  of  what  number? 

8  bu.  1  pk. 

17.  The  number  of  pupils  attending  school  on  a  cer- 
tain day  was  37;  this  w^as  7|%  less  than  the  number 
enrolled  :  how  many  w^ere  enrolled  ?  40. 

18.  A  gold  dollar  weighs  25.8  grains  Troy;  10%  of 
it  is  alloy:  how  many  grains  of  pure  gold  does  it  con- 
tain ?  23.22 


PERCENTAGE.  205 

19.  The  five-cent  piece  weiglis  5  grams,  of  whicli  1.25 
G.  are  nicl<:el  and  tlie  remainder  copj^er:  what  is  the  per 
cent  of  copper?  75%. 

20.  A  man  sold  a  horse  for  $150,  which  was  25%  more 
than  it  cost  him :  if  he  had  sold  the  horse  for  $200, 
how  many  per  cent  would  it  have  been  more  than  it 
cost  him?  66f%. 


APPLICATIONS  OF  PERCENTAGE. 


DEFINITIONS. 

170.  1.  The  applications  of  Percentage  may  be  di- 
vided into  two  classes:  (1)  Those  without  the  element 
of  time;   (2)  those  with  the  element  of  time. 

2.  The  most  important  applications  of  the  first  class 
are  (1)  Mercantile  Transactions,  (2)  Stock  Transactions. 

3.  The  most  important  applications  of  the  second  class 
are  (1)  Interest;   (2)  Discount. 

4.  Percentage  enters  to  a  greater  or  less  extent  into 
the  calculations  of  Exchange,  Insurance,  Taxes,  Equation 
of  Payments,  etc. 

5.  The  principles  of  Percentage  apply  directly  to  ap- 
plications ^of  ■  the  first  class  in  accordance  with  the  fol- 
lowing       pr"^    . 

Jyi/  GENERAL   RULE. 

V 

Rule. — 1.  Ascertain  the  quantities  which  correspond  to 
base,  percentage,   and   their  sum   or  their   difference. 

2.  Note  the  quantities  given  and  the  quantities  required. 

3.  Apply  the  proper  case  of  Percentage  to  the  given  ex- 
ample. 


206  KAYS  NEW  PKACTICAL  AKITHMKTIO. 

MERCANTILE  TRANSACTIONS. 
DEFINITIONS. 

171.  1.  Mercantile  Transactions  relate  to  the  pur- 
chase and  sale  of  merchandise. 

2.  Price  is  the  value  of  any  thing  in  mone}^ 

3.  Merchandise  is  bought  and  sold  at  wholesale  and  at 
retail  prices. 

4.  The  wholesale  price  is  the  price  of  merchandise  in 
quantities. 

5.  The  retail  price  is  the  price  of  merchandise  in 
small  quantities. 

Rem. —Wholesale  merchants  buy  and  sell  merchandise  at  whole- 
sale prices.  Retail  dealers  distribute  merchandise  of  every  descrip- 
tion, to  the  users  or  consumers  of  it,  at  retail  prices. 

6.  The  chief  Mercantile  Transactions  involving  an 
application  of  Percentage  are  (1)  Commission;  (2)  Trade 
Discount,  and  (3)  Pj^ofit  and  Loss. 

Rem. — Wholesale  merchants  buy  and  sell  merchandise  largely 
through  agents,  who  receive  salaries,  or  a  commission,  for  their  serv- 
ices; buyers  at  wholesale  are  sometimes  allowed  discounts  upon  their 
purchases;  and  merchants  usually  make  a  profit,  or  sufier  a  loss,  in 
their  transactions. 

COMMISSION. 

172.  1.  An  agent  is  a  person  intrusted  with  the 
business  of  another. 

Rem. — The  person  who  employs  the  agent,  in  reference  to  him,  is 
called  the  principal. 

2.  A  commission  merchant  buys  and  sells  merchan- 
dise for  another. 


COMMISSION.  207 

Kem.  1. — A  factor  is  an  agent  who  buys  and  sells  merchandise  in 
his  own  name,  and  is  intrusted  by  his  principal  with  the  possession 
and  control  of  it. 

Kem.  2. — The  person  to  whom  merchandise  is  sent  to  be  sold  is 
termed  the  consignee;  the  person  who  sends  it  is  termed  the  con- 
signor; while  the  merchandise  itself  is  called  a  consignment. 

3.  The  commission  is  the  sum  paid  an  agent  for 
transacting  business. 

4.  The  charges  are  expenses  incurred  by  an  agent  in 
transacting  business. 

5.  The  net  proceeds  is  the  sum  remaining  after  de- 
ducting  the  commission  and  charges. 

6.  The  value  of  the  materials  in  the  business  trans- 
acted is  the  base;  the  commission  is  the  percentage ;  and 
the  net  proceeds  is  the  base  less  the  percentage. 

Examples. 

1.  An  agent  whose  commission  is  5^,  receives  how 
much  upon  a  sale  of  goods  amounting  to  $240?  |12. 

2.  An  auctioneer  received  $11.50  for  selling  a  lot  of 
goods  amounting  to  $460 :  what  per  cent  commission 
did  he  receive?  2^%. 

3.  At  a  commission  of  2^^  a  commission  merchant 
receives  $8.12^  for  selling  25  barrels  of  molasses:  for 
how  much  per  barrel  did  he  sell  the  molasses?  $13. 

4.  An  agent  receives  $210  with  which  to  buy  goods : 
after  deducting  his  commission  of  5^  what  sum  must 
he  expend?  $200. 

5.  What  are  the  net  proceeds  on  a  sale  of  goods 
amounting  to  $180,  at  4%   commission?  $172.80. 

6.  A  lawyer  received  $11.25  for  collecting  a  debt:  his 
commission  being  5^,  what  was  the  amount  of  the 
debt?  $225. 


2(18  KAY'S  NEW  PKACTICAL  ARITHMETIC. 

7.  An  agent  receives  $1323.54  to  cover  cost  of  goods 
and  commission  at  8%  :  what  is  his  commission? 

$98.04. 

8.  A  commission  merchant  sells  250  bbl.  pork,  at  $15 
per  bbl.;  175  bbl.  flour,  at  $7  per  bbl.;  and  1456  lb. 
feathers,  at  25  ct.  per  lb.;  his  commission  is  3%:  what 
sum  does  he  remit  the  owner?  $5178.83. 


TRADE   DISCOUNT. 

173.  1.  Merchandise  may  be  sold  at  a  net  price  or  at 
a  discount  from  an  assumed  list,  or  regular,  price. 

2.  A  net  price  is  a  fixed  price  from  which  no  discount 
is  allowed. 

3.  A  list,  or  regular,  price  is  an  established  price,  as- 
sumed by  the  seller  as  a  basis  upon  which  to  calculate 
discounts. 

4.  The  discount  is  the  deduction  from  the  list,  or  reg- 
ular, price. 

Rem.  1. — In  the  wholesale  trade,  the  amount  of  discount  granted 
to  a  purchaser  depends  upon  ( 1 )  the  amount  purchased,  and  ( 2 )  the 
time  of  payment. 

Rem.  2. — In  some  lines  of  goods  the  discounts  are  made  from  the 
price-list  of  the  dealer;  in  others,  from  the  price-current  of  the 
market. 

Rem.  3. — In  regard  to  time,  selling  for  cash  means  payment  as 
soon  as  the  goods  can  be  delivered. 

Time  purchases  means  that  the  payments  are  to  be  made  in  a  cer- 
tain time  after  the  purchase — the  time  varies  with  different  lines  of 
goods. 

5.  The  discount  is  expressed  as  so  many  per  cent  off 
or  as  so  many  off. 

Thus,  20  (fc  off,  or  20  off,  means  at  a  discount  of  20  %  from  the 
price. 


TRADE  DISCOUNT.  209 

6.  There  may  be:  1st.  A  single  discount;  as  5^,  or  5 
off.     2d.  Two  or  more  successive  discounts. 

Thus,  the  expression  20  and  5  cL  off  means,  first,  a  discount  of 
20^  from  the  price,  and  then  a  discount  of  5^  from  the  remainder. 
The  expression  25,  10,  and  5  cL  oif,  means  three  successive  discounts. 

Rem.— The  per  cent  is  sometimes  expressed  as  a  common  fraction. 
Thus,  1  off  means  12^  cj^  off;  \  and  5  off  means  33i  and  5  off. 

7.  The  price  of  the  seller  is  the  base ;  the  sum  of  all 
the  discounts  is  the  percentage;  and  the  price  of  the 
buyer  or  price  paid  is  the  base  less  the  percentage. 


Examples. 

1.  A  bill  of  goods  amounted  to  $225.50  list;  20%  off 
being  allowed,  what  was  paid  for  the  goods?        $180.40. 

2.  A  bill  of  articles  amounted  to  $725.16,  the  purchaser 
being  allowed  i  and  5  off,  what  did  he  pay?       $459.27. 

3.  I  paid  $1430.75  for  a  lot  of  groceries,  which  was 
3%  discount  from  the  face  of  the  bill:  what  Avas  the 
amount  of  the  bill?  $1475. 

4.  A  bill  of  goods  cost  $390.45  at  25  and  5  off:  what 
was  the  list  price?  $548. 

5.  Sold  20  doz.  feather  dusters,  giving  the  purchaser 
a  discount  of  10,  10,  and  10%  :  his  discounts  amounting 
to  $325.20,  how  much  was  my  price  per  dozen?        $60. 

6.  Bought  100  doz.  stay  bindings,  at  60  ct.  per  dozen, 
for  40,  10,  and  7^^   off:    what  did  I  pay  for  them? 

$29.97. 

7.  A  retail  dealer  buys  a  case  of  slates  containing  10 
dozen  for  $50  list,  and  gets  off  50,  10,  and  10^  ;  paying 
for  them  in  the  usual  time,  he  gets  an  additional  2^  : 
what  did  he  pay  per  dozen  for  the  slates?  $1.98. 


210  HAY'S  NEW  PRACTICAL  ARITHMETIC, 


PROFIT  AND   LOSS. 

174.     1.  The  cost  is  the  i)rice  paid  for  goods. 

2.  The    selling    price    is  the  price  received    for  goods. 

Rem. — The  cost  to  the  consumer  is  the  selling  price  of  the  mer- 
chant; and  the  cost  to  the  retail  dealer  is  the  selling  price  of  the 
wholesale  dealer. 

3.  Goods  are  usually  sold  at  a  profit  or  at  a  loss. 

4.  The  profit  is  what  the  goods  sell  for  more  than 
they  cost. 

5.  The  loss  is  what  the  goods  sell  for  less  than  they 
cost. 

6.  The  cost  is  the  base;  the  profit  or  the  loss  is  the 
percentage;  and  the  selling  price  is  the  sum  or  the  differ- 
ence of  the  base  and  percentage. 

Examples. 

1.  A  merchant's  profit  on  a  piece  of  cloth  which  cost 
$40    is  10%  :  for  how  much  does  he  sell  it?  $44. 

2.  Prints  that  cost  6  ct.  a  yard  arc  sold  for  5  ct.  a 
yard:  what  is  the  per  cent  of  loss?  16f%. 

3.  A  grocer,  by  retailing  coflPee  at  27  cents  per  pound, 
gains  12^%  :  w^hat  did  it  cost  per  pound?  24  ct. 

4.  Selling  a  lot  of  goods  at  a  loss  of  4%,  the  loss  on 
the  entire  lot  was  $15.30:  what  did  the  goods  cost? 

$382.50. 

5.  To  make  a  profit  of  37^%,  at  what  price  must  a 
dry -goods  merchant  sell  shawls  that  cost  $8?  $11. 

6.  A  bookseller  sells  a  grammar  for  90  ct.  which  cost 
75  ct. :   w^hat  is  his  gain  per  cent?  20%. 

7.  What  is  the  cost  of  tea,  which,  when  sold  at  6J% 
profit  yields  a  profit  of  5  ct.  per  pound  ?  80  ct. 


PROFIT  AND  LOSS.  211 

8.  A  grocer  sells  apples  at  $4.75  per  barrel,  making  a 
profit  of  18f  %  :  what  was  the  cost?  $4. 

9.  Sold  silk  at  $1.35  per  yard,  and  lost  10^  :  at  what 
price  per  yard  would  I  have  sold  it  to  make  a  profit 
of  16f  %  ?  '  $1.75. 

10.  A  peddler  bought  a  stock  of  goods  for  $874,  and 
disposed  of  them  at  a  profit  of  25%  :  how  much  money 
did  he  make?  $218.50. 

11.  If  a  bookseller  makes  25  ct.  on  an  atlas,  which  he 
sells  for  $1.75:  what  is  his  per  cent  of  profit?         16f%. 

12.  A  dealer  sold  two  horses  for  $150  each  ;  on  one 
he  gained  25%,  and  on  the  other  he  lost  25%  :  how 
much  did  he  lose  by  the  transaction?  $20. 

13.  A  merchant  reduced  the  price  of  a  certain  piece 
of  cloth  5  ct.  per  yard,  and  thereby  reduced  his  profit 
on  the  cloth  from  10%  to  8%  :  what  was  the  cost  of 
the  cloth  per  yard?  $2.50. 

14.  A  speculator  bought  10000  bushels  of  corn,  at  60 
ct.  per  bushel;  in  a  few  days,  corn  advancing  in  price, 
he  sold  7000  bushels,  at  65  ct.  per  bushel  ;  corn  then 
falling  in  price,  he  disposed  of  the  remainder  at  55  ct! 
per  bushel :  what  per  cent  profit  did  he  get  out  of  the 
transaction?  '^i%. 

15.  A  speculator  in  real  estate  sold  a  house  and  lot 
for  $12000,  which  sale  afforded  him  a  profit  of  33J%  on 
the  cost;  he  then  invested  the  $12000  in  city  lots,  which 
he  was  obliged  to  sell  at  a  loss  of  33  J  %  :  how  much  did 
he  lose  by  the  two  transactions?  $1000. 

Miscellaneous  Examples. 

175.  1.  A  bookseller  purchases  books  from  the  pub- 
lisher at  20%  off  the  list  price;  if  he  retail  them  at  the 
list  price,  what  wnll  be  his  per  cent  of  profit?  25%. 


212  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  A  grocer  bought  5  luilf-chests  of  tea  of  74  lb.  each, 
at  45  ct.  per  lb.,  at  2^  off  for  cash :  if  he  retail  it  at 
12^^   advance,  what  will  be  his  profit?  ^20.12. 

8.  Bought  5  assorted  cases  of  men's  boots,  containing 
12  pairs  each,  for  $45  per  case,  5^  off  for  cash ;  I  retail 
them  at  $4.25  a  pair:  what  is  my  profit?  $41.25. 

4.  Sold  a  case  of  hats  containing  3  dozen,  on  which  1 
had  received  a  discount  of  10^  and  made  a  profit  of 
12^^  or  374  ct.  on  each  hat:  what  was  the  wholesale 
merchant's  j)rice  per  case?  $120. 

5.  A  merchant  bought  100  packs  of  pins,  of  12  papers 
each,  for  $1.00  per  pack,  00,  5  and  5%  off;  if  he  retail 
them  so  as  to  make  a  profit  of  $23.90,  lor  how  much  a 
paper  will  he  sell  them?  5  ct. 

6.  I  sent  a  car-load  of  flour,  100  bbl.,  to  a  commission 
merchant  in  New  York ;  he  disposed  of  the  flour  at 
$9.50  per  barrel,  his  commission  was  2-^%  with  charges 
of  $17.25:  if  the  flour  cost  me  $7.50  per  barrel,  how 
much   did   I  make?  $159. 

7.  A  contractor  bought  80  horses  for  government,  at 
$125  apiece;  the  freight  was  $200,  and  the  agent's  com- 
mission was  such  that  the  horses  cost  the  government 
$10450:  what  per  cent  was  the  commission?  2^%- 

8.  A  commission  merchant  sells  a  consignment  of  50 
hhd.  of  sugar,  1500  lb.  each  net,  at  10^  ct.  per  pound; 
his  commission  is  2%  and  charges  $22.50;  the  con- 
signor clears  14^  by  the  transaction:  what  did  he  pay 
per  pound  for  the  sugar?  9  ct. 

9.  A  dealer  in  notions  buys  60  gross  shoe-strings,  at  70 
^t.  per  gross,  list,  for  50,  10  and  5%  off;  if  he  sell  them 
at  20,  10  and  5%  off  list,  what  will  be  his  profit?     $10.77. 

10.  Bought  50  gross  of  rubber  buttons  for  25,  10  and 
5%  off;  disposed  of  the  lot  for  $35.91,  at  a  profit  of  12^  : 
what  was  the  list  price  of  the  buttons  per  gross?       $1.00. 


STOCK  TRANSACTIONS.  213 

STOCK  TRANSACTIONS. 
DEFINITIONS. 

17G.  1.  Stock  Transactions  relate  to  the  purchase 
and  sale  of  stocks,  bonds,  and  gold. 

2.  Stock  is  capital  in  the  form  of  transferable    shares. 

Rem. — The  capital  of  banks,  of  railroad,  insurance,  telegraph  and 
other  companies  is  held  in  this  way. 

3.  The  Stockholders  are  the  owners  of  the  stock. 

4.  A  share  is  usually  ^100. 

Rem. — A  share  is  sometimes  $50  or  some  other  number.  Stocks 
are  quoted,  in  the  New  York  market,  invariably  as  $100  to  the 
share. 

5.  A  bond  is  a  written  promise,  under  seal,  to  pay  a 
certain  sum  of  money  at  a  specified  time. 

Rem.  1. — Bonds  are  the  notes  of  the  Government  and  of  the  vari- 
ous corporate  bodies  which  are  allowed  to  issue  them;  usually  they 
bear  a  given  rate  of  interest  and  are  payable  within  a  specified  time. 

Rem.  2. — In  quoting  United  States  bonds,  the  different  issues  are 
distinguished,  1st.  By  the  rate  of  interest;  as  6's,  5's,  4J's,  4's;  2d.  By 
the  time  at  which  they  mature;  as  5-20's,  which  are  payable  in  20 
years,  but  may  be  paid  after  5  years.  The  5-20's  are  also  distin- 
guished by  the  date  of  their  issue,  as  5-20's  of  1868.  Bonds  of  the 
Funded  Loan  bear  5%  interest,  and  later  ones  4J  and  4^. 

Rem.  o. — The  bonds  of  local  corporations  take  the  name  of  the 
company  which  issues  them;  as,  "Chicago  and  Northwestern," 
"  Adams   Express,"   "  Western   Union   Telegraph,"  etc. 

6.  Currency  is  the  paper  money  of  the  country. 

Rem. — It  consists  of  legal-tender  notes,  called  "greenbacks,"  and 
National'  Bank   notes. 


214  KArS  KEW  PRACTICAL  ARITHMETIC. 

7.  The  par  value  of  stocks  and  bonds  is  the  value 
given  on  the  face  of  them. 

Hem. — The  quotations  for  stocks,  bonds,  and  gold  are  all  based  on 
the  currency  dollar. 

8.  The  chief  Stock  transactions  involving  an  applica- 
tion of  Percentage  are  (1)  Brokerage^  (2)  Assessments  and 
Dividends^  (3)  Stock   Values^  and  (4)  Stock  Investments. 


BROKERAGE. 

177.  1.  A  broker  is  an  agent  who  buys  and  sells 
stocks,   bonds,  gold,  etc. 

Rem. — Persons  who  "operate"  in  stocks  usually  do  so  through 
broken^;  the  latter  buy  and  sell  stocks  in  kind  and  amount  as  they 
are  authorized  by  the  "operator." 

2.  Brokerage  is  the  sum  paid  the  broker  for  transact- 
ing the  business,  and  is  calculated  on  the  par  value. 

3.  The  par  value  is  the  hase^  the  brokerage  the  2>^r- 
centage. 

Examples. 

1.  A  broker  bought  for  me  75  shares  New  York  Cen- 
tral and  Hudson  Eiver  stock :  required  the  brokerage  at 
J%.  $18.75. 

2.  The  brokerage  for  buying  50  shares  of  Chicago 
and  Eock  Island  stock  was  $6.25 :  what  was  the  per 
cent?  1^. 

3.  At  \%  brokerage  a  broker  received  $10  for  making 
an  investment  in  bank  stock  :  how  many  shares  did  he 
buy?  40. 


ASSESSMENTS  AND  DIVIDENDS.  215 

4.  A  broker  buys  17  shares  Milwaukee  and  St.  Paul 
preferred  stock :  what  is  his  brokerage,  at  J^  ?       $4.25. 

5.  The  brokerage  on  95  shares  of  Vermont  Central 
stock  is  $11. 87^:  what  is  the  per  cent?  -J^. 

6.  A  broker  received  $9.50,  or  a  brokerage  of  ^%,  for 
buying  Union  Pacific  stock :  how  many  shares  did  he 
purchase  ?  38. 

ASSESSMENTS  AND  DIVIDENDS. 

178.  1.  An  assessment  is  a  sum  of  money  paid  by 
the  stockholders. 

Kem. — In  the  formation  of  a  company  for  the  transaction  of  any 
business,  the  stock  subscribed  is  not  usually  all  paid  for  at  once;  but 
assessments  are  made  from  time  to  time  as  the  needs  of  the  business 
require.     The  stock  is  then  said  to  be  paid  for  in  installments. 

2.  A  dividend  is  a  sum  of  money  paid  to  the  stock- 
holders. 

Rem. — The  gross  earnings  of  a  company  are  its  total  receipts  in  the 
transaction  of  the  business;  the  net  earnings  are  what  is  left  of  the 
receipts  after  deducting  all  expenses.  The  dividends  are  paid  out 
of  the  net  earnings. 

Examples. 

1.  I  own  35  shares  of  bank  stock;  if  the  bank  de- 
clare a  dividend  of  4^,  what  will  I  receive?  $140. 

2.  A  man  pays  an  assessment  of  7^%,  or  $300,  on  his 
insurance  stock:  how  many  shares  does  he  own?         40. 

3.  A  mining  company  declares  a  dividend  of  15^  : 
what  does  Mr.  Jones  receive  who  owns  80  shares  of 
stock?  $1200. 

4.  A  man  owns  60  shares  of  railroad  stock :  if  the 
company  declare  a  dividend  of  5^  payable  in  stock,  how 
much  stock  will  he  then  own?  03  shares. 


216  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

5.  A  gas  company  has  a  capital  stock  of  $lt)0000;  its 
gross  earnings  are  $15700,  and  its  expenses  $4500  annu- 
ally:  what  per  cent  does  it  pay  the  stockholders?      1%- 

STOCK   VALUES. 

179,  1.  The  market  value  of  stocks,  bunds,  and  gold 
is  the  price  at  which  they  sell. 

Rem. — Stock  is  above  pnr,  or  at  a  premium,  when  it  sells  for  more 
than  the  par  value;  stock  is  below  jxir,  or  at  a  disconni,  when  it  sells 
for  less  than  the  par  value. 

2.  The  market  value  of  stocks,  bonds,  and  gold  is 
estimated  at  a  certain  per  cent  of  the  ])ar  value. 

Thus,  "gold,  106^,"  means  that  the  gold  dollar  is  worth  106J^  of 
the  currency  dollar,  or  is  at  a  premium  of  6|  ^.  "  New  York  Cen- 
tral and  Hudson  River,  91  J,"  means  that  the  stock  of  this  railroad 
sells  for  91^  ol  of  the  par  value,  or  is  at  a  discount  of  Sh  4). 

3.  The  par  value  is  the  base;  the  premium  or  discount  is 
the  percentage;  the  market  value,  the  amount  or  difference. 

Examples. 

1.  What  will  be  the  cost  of  150  shares  ($50  each)  of 
Harlem,  at  139|,  brokerage  \%  ?  $10500. 

2.  Bought  $8000  in  gold  at  110,  brokerage  ^^  :  what 
did  I  pay  for  the  gold  in  currency?  $8810 

3.  My  broker  sells  50  shares  of  Chicago  and  North 
western,  brokerage  ^%  ;  he  remits  me  $2475 :  at  what 
per  cent  did  the  stock  sell?  49f^. 

4.  What  will  be  the  cost  of  25  1000-dollar  5-20  U.  S. 
Bonds  of  1867,  at  114J,  brokerage  i%  ?  $28593.75. 

5.  I  paid  $1560  for  Milwaukee  &  St.  Paul,  at  19i 
brokerage  \%  :  how  many  shares  did  I  buy?  80. 


STOCK  INVESTMENTS.  217 

6.  When  gold  is  at  105,  what  is  the  value  in  gold  of 
a  dollar  in  currency?  ^^2T  ^^' 

7.  When  gold  was  at  112J,  what  was  the  value  of  a 
dollar  in   currency?  88|-  ct. 

8.  In  1864,  the  "greenback"  doilar  was  worth  only 
35f  ct.  in  gold  :  what  wa^  the  price  of  gold  ?  280. 

9.  A  merchant  paid  $8946.25  for  gold,  at  105,  broker- 
age \^c  •  l^<)w  much  gold  did  he  buy?  $8500. 

10.  My  broker  sells  a  certain  amount  of  gold,  and 
remits  me  $25734.37^?  His  brokerage,  at  xV%'  ^^® 
$15,621:  what  was  the  price  of  the  gold?  103. 

STOCK    INVESTMENTS. 

180.  1.  The  income  is  the  annual  profit  from  the 
investment. 

Eem;— The  income  from  most  of  the  United  States  bonds  is  in 
coin  or  its  equivalent. 

2.  The  cost  of  the  investment  is  the  base;  the  income 
is  the  'percentage. 

Examples. 

1.  If  I  invest  $39900  in  6^  bonds,  at  par,  what  will 
be  my  income?  $^394. 

2.  If  I  invest  $39900  in  6%  bonds,  at  105,  what  will 
be  my  income?  $2280. 

3.  if  I  invest  $39900  in  6%  bonds,  at  95,  what  will  be 
my  income?  $2520. 

4.  What  is  a  man's  income  who  owns  20  1000-dollar 
IT.  S.  6^  bonds,  when  gold  is  107?  $1284. 

5.  What  income  in  currency  would  a  man  receive  by 
investing  $5220  in  U.  8.  5-20  6^  bonds,  at  116,  when 
gold  is  ^105?  ^283.50. 


218  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

6.  What  per  cent  of  income  do  U.  S.  4 J  per  cents,  at 
108,  yield  when  gold  is  105?  4g%. 

7.  If  I  receive  an  annual  dividend  of  G^  on  Michigan 
Central  stock,  which  cost  me  but  37^,  what  per  cent  of 
income  do  I  receive  on  my  investment?  16^. 

8.  What  sum  invested  in  U.  S.  S's  of  1881,  at  118, 
yielded  an  annual  income  of  $1921  in  currency,  when 
gold  was  at  113?  $40120. 

9.  How  many  shares  of  stock  bought  at  95J,  and  sold 
at  105,  brokerage  \%  on  each  transaction,  will  yield  a 
profit  of  $925?  "^  100. 

10.  What  must  be  paid  for  6^  bonds  to  realize  an 
income  of  8^^?  75^. 

11.  When  U.  S.  4%  bonds  are  quoted  at  iOG,  what 
yearly  income  will  be  received  in  gold  from  the  bonds 
that  can  be  bought  for  $4982?  $188. 

12.  If  I  pay  87 J  for  railroad  bonds  that  yield  an  an- 
nual income  of  7^,  w^hat  per  cent  do  I  get  on  my  in- 
vestment? %%. 

13.  What  could  I  afford  to  pay  for  bonds  yielding  an 
annual  income  of  7^  to  invest  my  money  so  as  to 
realize  6%?  116f. 


DEFINITIONS. 

181.     1.  Interest  is  money  paid  for  the  use  of  money. 
Rem. — The  interest  is  paid  by  the  borrower  to  the  lender. 

2.  The  principal  is  the  money  for  the  use  of  which 
interest  is  paid. 

3.  The  amount  is  the  sum  of  the  principal  and  in- 
terest. 

4.  A  promissory  note  is  a  written  promise  to  pay  a 
certain  sum  of  money  at  a  specified  time. 

Rem. — The  borrower  always  gives  the  lender  his  note  for  the 
money.     The  following  is  a  common  form: 

$500.00.  Dayton,  O.,  June  IG,  1877. 

One  year  after  date  I  promise  to  pay  Clu^*les  Thomas, 
or  order,  five  hundred  dollars,  with  interest  at  8^,,  for 
value  received.  James  Q.  Dean. 

Rem. — "When  a  note  is  made  to  draw  interest  from  date,  the  words 
"from  date"  are  frequently  inserted  after  the  word  "interest." 

5.  The  face  of  the  note  is  the  principal. 

6.  Legal  interest  is  interest  at  a  per  cent  that  is 
allowed  b}^  law. 

(219) 


220 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


Rem. — The  per  cent  of  interest  that  is  legal  in  the  different  States 
and  Territories,  is  exhibited  in  the  following 


TABLE. 


NAME  OF  STATE. 


Alabama 

Arizona 

Arkansas 

California  

Colorado 

Connecticut 

Dakota 

Delaware 

District  Columbia. 

Florida  

Georgia 

Idaho 

Illinois    

Indiana  

Iowa 

Kansas 

Kentucky  

Louisiana 

Maine  -^ 

Maryland 

Massachusetts 

Michigan 

Minnesota 

Mississippi 


8^« 
6% 

6^ 

7<fo 
lOfc 

7</o 
6^ 

7fc 


Any. 

Any. 
Anv. 


12  fo 


Any. 
Any. 

lOfc 

10% 

12/. 

Any. 


lOfc 
12/, 
10% 


NAME  OF  STATE. 


MiBSouri    

Montana 

Nebraska 

Nevada 

New  Hampshire... 

New  Jersey 

New  Mexico 

New  York 

North  Carolina 

Ohio 

Oregon 

Pennsylvania 

Rhode  Island 

South  Carolina 

Tennessee 

Texas 

United  States 

Utah 

Vermont  .."T. 

Virginia 

Wash.  Territory... 

West  Virginia 

Wisconsin 

Wyoming  


60/, 
lOf, 
10/. 
10/. 

6/. 

6/. 

6/. 


6/f 


6/. 

7/. 

6/. 

8/. 

6/. 
10/. 

6/. 

0/. 
10/. 

6/. 

7/. 
12/. 


10/. 
Any. 
12/. 
Any. 


12/. 


8/. 
12/. 


Any. 
Any. 


12/. 
Any. 


Any. 


10/. 
Anv. 


When  the  per  cent  of  interest  is  not  mentioned  in  the  note  or  con- 
tract, the  first  column  gives  the  per  cent  that  may  be  collected  by 
law.  If  stipulated  in  the  note,  a  per  cent  of  interest  as  high  as  that 
in  the  second  column  mav  be  collected. 


SIMPLE  INTEREST.  221 

7.  Usury  is  charging  interest  at  a  per  cent  greater 
than  that  allowed  by  law. 

Rem. — It  will  be  seen  from  the  table  above  that  usury  is  now 
practically  abolished  in  nearly  half  the  States  and  Territories. 

8.  The  subject  of  Interest  may  be  divided  into  (1) 
Simple  Interest^  (2)  Compound  Interest^  (3)  Annual  Inter  est , 
(4)  Partial  Payments. 


SIMPLE   INTEREST. 

182.  1.  Simple  Interest  is  interest  on  the  principal 
onl}^ 

Rem. — Simple  interest  is  not  due  and  can  not  be  collected  till  the 
principal  is  due. 

2.  In  Simple  Interest  four  quantities  arc  considered, 
(1)  the  principjal,  (2)  the  per  cent,  (3)  the  tirne,  and  (4) 
the  interest. 

3.  Any  three  of  these  quantities  being  given,  the  fourth 
may  be  found.     There  are  Jive  cases. 

CASE    r. 

183.  Given  the  principal,  the  time  and  the  per  cent, 
to  find  the  interest. 

1st.     When  the  time  is  one  7jear, 
1.  Find  the  interest  of  $25  for  1  yr.,  at  6%. 

OPKnATlON. 

SoLtJTioN.— 6  ^^   is  !f)6  (Art.  162,  3).     Then,  since  2  5 

one  year  is  the  unit  of  time,  the  interest   for  1  yr.  is  .0  6 

$25  X. 06  ^.$1.50.  IJO 


222  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  Find  the  interest  of  818.75  for  1  yr.,  at  6§%. 

OPKRATION. 

15)18.75(1.25 
15 
Solution.— 6|^o  is  ^,  (Art.  162,  3,  Rem.  3  7 

2).     Then,  the  interest  for  1  yr.  is  $18.75 -i-  3  0 

15  r=  $1.25.  7  5 


V  b 


3.  Find  the  amount  of  S215  for  1  yr.,  at  G%. 

OPERATION. 
$2  15 

Solution.— The  interest  of  $215  for  1  yr.  ni  6  fo  .0  6 

U    $12.90;     then,    the    amount    is    $2\ry -\- $V2/J0  ^  1  2.9  0 

$227.90.  2  1  5 

$  2  2  7.9  0 

Hule. — Multiply  the  principal  by  the  rate. 

Rem. — To  find  the  amount  add  the  principal  and  interest. 

Find  the  interest 

V 

4.  Of    $200  for  1  yr.,  at  S%.  $16.00. 

5.  Of    $150  for  1  yr.,  at  5^.  $7.50. 

6.  Of      $85  for  1  yr.,  at  7%.  $5.95. 

7.  Of  $7200  for  1  yr.,  at  GJ^.  $450. 

Find  the  amount 

8.  Of  $28.20  for  1  yr.,  at  8J%.  $30.55. 

9.  Of  $45.50  for  1  yr.,  at  10%.  $50.05. 

10.  Of  $420  for  1  yr.,  at  ^%.  $442.40. 

11.  Of  $857  for  1  yr.,  at  9%.  $934.13. 

12.  Of  $96  for  1  yr.,  at  8^%.  $104.16. 

13.  Of  $2000  for  1  yr.,  at  U^.  $2090. 

14.  Of  $164  for  1  yr..  at  m%.  $184.50. 


SIMPLE    INTEKEST.  223 

2d.     When  the  time  is  Two  or  More  Years. 

1.  Find  the  interest  of  $50  for  3  yr.,  at  7%. 

SonjTioN. — The  interest  of  $50  for  1  yr.,  at  7^,  is  operation. 

$3.50;    then,   the   interest   for   3   yr.   is   $3.50X3=  $^0 

$10.50.  .0  7 

3.5  0 

Rem. — It   is   sometimes   more   convenient    first   to  3 

multiply  the   per   cent   and   time   together.     In   the  $  1  0.5  0 
above  example,  the  per  cent  for  3  yr.  is  21. 

2.  Find  the  amount  of  $225.18  for  3  yr.,  at  4^%. 

OPERATION. 

Solution.— The  interest  of  $225.18  for  \  yr.,  at  $  2  2  5.1  8 

A\c/ci  is  $10.1331;    then,   the  interest   for  3  yr.  is  ^0  4| 

$10.1331  X  S  =  $30.3993;       and     the     amount     is  9  0  0  7  2 

$30.3993  -I-  $225.18  ^  $255.58.  112  5  9 

roX3~3T 

Rem. — In  business,  it  is  customary  to   take  the  3 

final  result  to  the  nearest  unit.     Thus,  in  the  ex-  3  0.39  9  3 

ample,  57  cents  9  mills  and  3  tenths  of  a  mill  are  2  2  5.1  8 

nearest  58  cents.  $  2  5  5.5  7  9  3 

Rule. — 1.  Find  the  interest  for  one  year.     3Iultiply  this 
by  the  given  number  of  years. 

Find  the  interest 

3.  Of  S65  for  4  yr.,  at  5%.  $13. 

4.  Of  $300  for  2  yr.,  at  6%.  $36. 

5.  Of  $275  for  3  yr.,  at  6%.  $49.50. 

6.  Of  $187.50  for  4  yr.,  at  5%.  $37.50. 

7.  Of  $233.80  for  10  yr.,  at  6%.  $140.28. 

Find  the  amount 

8.  Of  $45  for  2  yr.,  at  8%.  $52.20. 

9.  Of  $80  for  4  yr.,  at  7%.  $102.40. 


224 


KAY'S  NEW   PRACTICAL  ARITHMETIC. 


10.  Of  $237.16  for  2  yr.,  at  3f  %. 

11.  Of  $74.75  for  5  yr.,  at  4%. 

12.  Of  $85.45  lor  4  yr.,  at  6%. 

13.  Of  $325  for  3  yr.,  at  5|%. 

14.  Of  $129.36  for  4  yr.,  at  4f%. 

15.  Of  $8745  for  2  yr.,  at  8%. 


$254.95. 

$89.70. 

$105.96. 

$377.65. 

$152. 

$10144.20. 


3d,     When  the  Time  is  atiy  J^iunhev  of  Months. 
1.  Find  the  interest  of  $24  for  9  nio.,  at  6%. 


OPBRATION. 


Solution  I. — 9  mo.  are  ij  of  ji  year.  The  interest  of 
$24  for  1  yr.,  at  6%,  i^  $1.44;  then,  the  interest  for  9 
mo.  is  J  of  $1.44,  which  is  $1.08. 


Solution  II.— (Art.  130)6  mo.  arc  h  of 

u  year,  and  3  mo.  are  \  of  6  mo.  The  inter- 
est of  $24  for  1  yr.,  at  65^,  is  $1.44;  then,  the 
interest  for  6  mo.  is  \  of  $1.41,  whieh  is  72 
ct.,  and  the  interest  for  8  mo.  is  \  of  72  et.,  6  mo. 

which  is  36  ct.     Then,  the  interest  for  9  mo.  3  mo. 

is  72  ct.  +  36  ct.  =r$1.08. 


Rule. — 1.  Find  the  interest  for  one  year.  Take  such  a 
part  of  this  as  the  given  number  of  7honths  is  part  of  a 
year. 


Find  the  interest 


2.  Of  $300  for 

3.  Of  $240  for 

4.  Of  $  50  for 

5.  Of  $  86  for 

6.  Of  $  50  for 


1  mo., 

2  mo., 
5  mo., 

3  mo., 

4  mo., 


at  6%. 
at  S%. 
at  6%. 
at  6%. 
at  8%. 


$1.50. 
$3.20. 
$1.25. 
$1.29. 
$1.33. 


SIMPLE  INTEREST. 


225 


Find  the  amount 
7.  Of  S150.25  for 
Of  S360        for 

Of  $204  for 
Of  $228  for 
Of  $137.50  for 


at  8%. 
at  5%. 
at  7^. 
at  6%. 
at  6%. 
12.  Of  S759()      for  10  mo.,  at  8%. 


8. 

9. 
10. 
11. 


G  mo., 

7  mo., 
11  mo., 

9  mo., 

8  mo., 


$156.26, 
$370.50. 
$217.09. 

$238.26. 

$143.00. 

$8102.40. 


^th.     When  the  Time  is  any  JWoinher  of  Days, 
1.  Find  the  interest  of  $288  for  24  da.,  at  5^. 


Solution  I.— 24  da.  are  4  of  a  month.  The  in- 
terest of  $288  for  1  mo.,  at  5^^^  is  $1.20;  then,  the 
interest  for  24  da.  is  |  of  $1.20,  which  is  90  ct. 


OPERATION. 

•288 
.0  5 


Solution  II.— (Art.  130)  15  da.  are  h 
of  a  month,  6  da.  are  \  of  a  month,  and  3 
da.  are  ^  of  6  da.  The  interest  of  $288  for 
1  mo.  at  5^,  is  $1.20;  then,  the  interest  for 
15  da.  is  I  of  $1.20,  which  is  60  ct.;  the  in- 
terest for  6  da.  is  -i  of  $1 .20,  which  is  24  ct., 
and  the  interest  for  3  da.  is  I  of  24  ct., 
which  is  12  ct.  Then,  the  interest  for  24 
da.  is  60  ct.  +  24  ct.  -[-  12  ct.  —  ^Q  ct. 


1  2  )  1  4.4  0 

5  )  1.2  0 

.2  4 

_4 

.9li 


OPERATION. 

288 
.0  5 
1  2  )1  4:40 


1.2  0 

15  da. 
6  da. 
3  da. 

.6  0 
.2  4 
.12 

.9  6 


Rule. — 1.  Pind  the  interest  for  one  month.  Take  such  a 
part  of  this  as  the  given  number  of  days  is  part  of  a 
month. 


Rem. — In  computing  interest,  it  is  customary  to  regard  80  days  as 
1  month. 

Prac.  15. 


22G  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

Find  the  interest 

2.  Of  S360    for  20  dii.,  at  6%.  $1.20, 

3.  Of  S726     for  10  dti.,  tit  6^.  $1.21 

4.  Of  $1200  for  15  da.,  at  6^.  $3.00 

5.  Of  $180     for  19  da.,  at  8%.  76  ct 

6.  Of  $240    for  27  da.,  at  7%.  $1.26 

7.  Of  $320     for  21  da.,  at  5%.  93  ct. 

8.  Of  $450     for  25  da.,  at  10%.  $3.13. 

Find  the  amount 

9.  Of  $100.80  for  28  da.,  at  5%.  $101.19. 

10.  Of  $150       for  18  da.,  at  5%.  $150.38. 

11.  Of  $360        for  11  da.,  at  6^.  $360.66. 

12.  Of  $264        for     9  da.,  at  6%.  $264.40. 

13.  Of  $900        for  14  da.,  at  7^.  $902.45. 

14.  Of  $430        for  19  da.,  at  4^%.  $431.02. 


Sth.  When  the  Time  is  Tears,  Months,  and  Days, 
or  any  Two  of  these  Periods. 

First  Method. 

1.  Find    the   interest   of  $360    for  2  yr.  7  mo.  25  da., 
at  8%. 

OPERATION. 

$360  1  2  )  2  8.8  0 

Solution  I.— The  interest  of  $360  for                 .0  8  2.4  0 

1  yr.,  at  8^^,  is  $28.80;   then,  for  2  yr.  2  8.8  0  7_ 

the  interest  is  $57.60;   for  7  mo.,  or  /^                     2  $16.8  0 
of  a  year,  the  interest  is  $16.80;  and  for            5  7.6  0 

25  da.,  or  f  of  a  month,  the  interest  is            1  6.8  0  6  )  2.4  0 

$2.     Then,  the  interest  for  2  yr.  7  mo.                2.0  0  M) 

25  da.  is  $57.60  +  $16.80  +  $2  =  $76.40.         $7  6.4  0  5 

$2.0  0 


simple:  INTEKE8T. 


227 


Solution  II.— (Art.  130).  The  inter- 
est of  $360  for  1  yr.,  at  8^^,  is  28.80,  and  for 
2  yr.  it  is  $57.60;  for  6  mo.,  or  half  of  a 
year,  the  interest  is  $14.40,  and  for  1  mo., 
or  ^  of  6  mo.,  it  is  $2.40;  for  15  da.,  or  ^  of 
a  month,  the  interest  is  $1.20,  and  for  10 
da.,  or  ^  of  a  month,  it  is  80  ct.  Then,  the 
interest  for  2  yr.  7  mo.  25  da.  is  $57.60  -f 
$14.40  +  $2.40  +  $1.20  +  $0.80  =  $76.40. 


OPERATION. 

$? 


6  mo.  =  ^ 

1  mo.  =  I 

15  da.  =:  ^ 

10  da.  =1 


$7 


7.6  0 
4.4  0 
2.4  0 
1.2  0 
.8  0 
6.4  0 


Rule  I.- 

the  results. 


-1.  Fhid  the  interest  for  each  period,  and  add 


Second   Method. 


/^ 


2.  Find  the  interest  of  $120    for    4  yr.    6  mo.    20  da., 
at  6%. 


Solution. — 20  da.  are  |  of  a  mo.; 
6|  mo.  are  |  of  a  year.  Then,  the  in- 
terest of  $120  for  4  yi-.  6  mo.  20  da., 
at  6^c,  will  be  $120  X -06X41  = 
$32.80. 


OPERATION. 


40  .02 


3  0  —   S 
Q2  2_0. 


32.80 


Rule  II. — 1 .   Reduce  the  months  and  days  to  the  fraction 
of  a  year. 

2.  Midtiply  the  principal   by   the   rate,  and  multiply   the 
product  by  the  time  expressed  in  years. 

Kem. — Indicate  the  operation  as  far  as  is  practicable,  and  employ 
cancellation. 

3.  Find  the  interest  of  $150  for  4  yr.  2  mo.,  at  6^. 

$37.50. 


228  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

Find  the  interest  of 

4.  S375.40  for  1  yr.  8  mo.,  fit  G^.  S37.54. 

5.  $  92.75  for  3  yr.  5  mo.,  at  6^.  $19.01. 

6.  $500  for  1  yr.  1  mo.  18  da.,  at  6^.  $34.00. 

7.  $560  for  2  yr.  4  mo.  15  da.,  at  8%.  $106.40. 

8.  $750  for  4  yr.  3  mo.     6  da.,  at  6^.  $192.00. 

9.  $456  for  3  yr.  5  mo.  18  da.,  at  5%.  $79.04. 

10.  $216  for  5  yr.  7  mo.  27  da.,  at  10^.  $122.22. 

11.  $380  for  3  yr.  9  mo.     9  da.,  at  15%.  $215.18. 

Find  tlie  amount  of 

12.  $300  for  3  yr.  8  mo.,  at  6%.  $366.00. 

13.  $250  for  1  yr.  7  mo.,  at  6%.  $273.75. 

14.  $205.25  for  2  yr.  8  mo.  15  da.,  at  6%.  $238.60. 

15.  $150.62  for  3  yr.  5  mo.  12  da.,  at  5^.  $176.60. 

16.  $210.25  for  2  yr.  7  mo.  20  da.,  at  7%.  $249.09. 

17.  $  57.85  for  2  yr.  3  mo.  23  da.,  at  5%.  $64.54. 

18.  Find  the  interest    of  $150,  from    January  9,  1847, 
to  April  19,  1849,  at  6%.  $20.50. 

Rem. — To  find  the  time  between  two  dates,  see  Art.  77. 

19.  The  interest  of  $240,  from    February    15,  1848,  to 
April  27,  1849,  at  8^.  $23.04. 

20.  The  interest  of  $180,  from  May  14,  1843,  to  August 
28,  1845,  at  7^.  $28.84. 

21.  The  interest  of  $137.50,  from  July  3,  to  November 

27,  at  9^.  $4.95. 

22.  The  amount  of  $125.40,  from  March    1,  to  August 

28,  at  Si%.  $130.64. 

23.  The  amount   of  $234.60,   from    August    2,  1847,  to 
March  9,  1848,  at  5J^.  $242.02. 

24.  The  amount  of  $153.80,  from  October  25,  1846,  to 
July  24,  1847,  at  5%.  $159.55. 


SIMPLE  INTEREST.  229 

184.  The  twelve  per  cent  method  of  finding  interest. 
Ist.  To  find  the  interest  of  SI  for  any  time,  at  12%, 

Explanation. — The  interest  of  $1  for  1  mo.,  at  \2(ij,  is  $0.01,  or 
1  ct.;  for  2  mo.,  it  is  2  ct.;  for  3  mo.,  it  is  3  et.,  etc.     Hence, 

The  ifiterest  of  $1  for  any  number  of  months,  at  VloL,  is  as  many 
cents  as  there  are  mofiths. 

The  interest  of  $1  for  3  da.,  at  12^/^,  is  $0,001,  or  1  mill;  for  6  da., 
it  is  2  mills;  for  9  da.,  it  is  3  mills,  etc.     Hence, 

The  interest  of  $1  for  any  number  of  days,  at  12^,  is  J  as  yna.ny 
mills  as  there  are  days. 

Rule. —  Call  the  months  cents,  and,  one-third  of  the  days 
mills. 

Rem. — Reduce  years  to  months. 

Find  the  interest  of  $1,  at  12^, 

1.  For  9  mo.  12  da.  $0,094 

2.  For  4  mo.  18  da.  .                                        $0,046, 

3.  For  7  mo.  12  da.  $0,074 

4.  For  9  mo.  3  da.  $0,091 

5.  For  1  yr.  4  mo.  $0.16, 

6.  For  1  yr.  5  mo.  27  da.                                  $0,179, 

7.  For  2  yr.  3  mo.  21  da.                                  $0,277 

8.  For  3  yr.  7  mo.  12  da.                                  $0,434 

9.  For  4  yr.  2  mo.  15  da.                                  $0,505 

10.  For  2  mo.     1  da.  $0.020J 

11.  For  5  mo.  17  da.  $0.055f 

12.  For  10  mo.  13  da.  $0.104J 

13.  For  1  yr.  2  mo.  4  da.  $0.141J 
14  For  2  yr.  9  mo.  20  da.  $0.336§ 
15.  For  3  yr.       5  mo.  29  da.  $0.419§ 


230 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


2d.    To    find  the  interest  of  $1,  for    any  time    at  any 
per  cent. 

1.  Find  the  interest   of  SI,  for  2  yr.  5  mo.  18  da.,  at 

Solution. — 6q^  is  i  of  12^.    The  interest  of  $1  operation. 
for  2  yr.  5  mo.  18  da.,  at  12^^,  is  $0,296;  then,  the  in-  2  ).2  9  G 

terest,  at  69^,  will  he  ^  of  $0,296,  which  is  $0,148.  ~ATs' 

2.  Find  the  interest  of  $1,  for   3  yr.  7  mo.  20  da.,  at 


operation. 
Solution.— 8^0  is  ^  of  Ufc    The  interest  of  $1  for        3)  .4  3  6^ 
3  yr.  7  mo.  20  da.,  at  12^^,  is  $0.436§;  then,  the  inter-  TTSJ 

est,  at  Sfo,  will  be  f  of  $0.43G§,  which  is  $0,291^.  2 

.29lj 

Rule.  -1.  Find   the  interest,  at   12^,  and  take  such  a 
fart  of  this  as  the  given  per  cent  is  of  12%. 


Find  the  interest  of  81, 


For  7  mo.  24  da.,  at  6%. 
For  10  mo.  15  da.,  at  5%. 
For  11  mo.  18  da.,  at  9^. 
2  mo.     9  da.. 


For  1 
For  2 
For  3 
For  4 


10.  For  5 


yr- 
yi*- 
yi^ 
yi'- 


5  mo. 

10  mo. 

3  mo. 

7  mo. 


12  da., 

17  da., 

11  da., 

24  da.. 


at 
at 
at 
at 
at 


10^. 
4^. 


$0,039. 
?0.043f. 

$0,087. 
$0.07  U. 

$0.1 9G. 

$0.388yV 

$0.299|f. 

$0,226. 


3d.  To  find  the  interest  of  any  sum    for  any  time,  at 
any  per  cent. 


SIMPLE  INTEREST.  231 

1.  Find  the  interest  of  S25,  for  1  yr.  5  mo.  18  da.,  at 


Solution. — The  interest  of  $1  for  1  yr.  5  mo.  18 
da.,  at  12t/c,  is  $0,176;  then,  at  6%,  it  is  $0,088. 
Then,  the  interest  of  $25  will  be  $0,088  X  25  ==:  $2.20. 


OPERATION. 

2).17  6 

.0  8  8 

25 

440 

176 

$2.2  0 

2.  Find  the  interest  of  $134.45,  for  1  yr.  7  mo.  15  da., 
at  S%' 

OPERATION. 

Solution.— The   interest    of   $1    for   1  3  ).l  9  5         $13  4.45 

yr.  7  mo.  15  da.,  at  1^^^.,  is  $0,195;  then,  .0  6  5 J_3 

at  Sfc,  it  is  $0.13.     Then,  the  interest  of  2  403  35 

$134.45  will  be  $.13  X  134.45  =  $17.48.  .13  0         1  3445 

$  1  7.4  7  8  5 

Rule. — 1.   Fhid   the  interest  of  $1,  and  multiply  this  by 
the  given  sum. 

Rem. — Take  either  f^ictor  for  the  multiplier  as  is  most  convenient. 

Find  the  interest 

3.  Of  $40,     for  6     mo.  21  da.,  at  6%.  $1.34 

4.  Of  $50,     for  8     mo.  24  da.,  at  9^.  $3.30 

5.  Of  $120,  for  10  mo.  12  da.,  at  7%.  $7.28 

6.  Of  $200,  for  11  mo.   15  da.,  at  6%.  $11.50 

7.  Of  $500,  for  1  yr.  3  mo.  G  da.,  at  3%.  $19 

8.  Of  $750,  for  1  yr.  5  mo.  27  da.,  at  S%.  $89.50 

9.  Of  $48.75,  for  1  yr.  9  mo.  3  da.,  at  6%.  $5.14 

10.  Of  $7G.32,  for  1  yr.  10  mo.  25  da.,  at  4^.     $5.81 

Find  the  amount 

11.  Of  $600,  for  2  yr.  1  mo.  9  da.,  at  5%.        $663.25. 

12.  Of  $900,  for  2  yr.  4  mo.   10  da.,  at  6%.    $1027.50. 


232  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

13.  Of  S86.25,  for  2  yr.  7  mo.  17  da.,  at  9%.     $10G.G7. 

14.  Of  S450,  for  3  yr.  2  mo.  13  da.,  at  8%.     $565.30. 

15.  Of  $534.78,  for  3  yr.  5  mo.  22  da.,  at  4%. 

S609.17. 

16.  Of  S1200,  for  3  yr.  11  mo.  15  da.,  at  10%.      $1675. 

CASE    II. 

185.     Given    the    principal,  the    jkm*    cent   and  the  in- 
terest, to  find  the  time. 

1.  The  interest  of  $225  for  a  certain  time,  at  4%,  was 
$66:  what  was  the  time? 

OPERATION. 

Solution.— The  interest  of  $225  fori  $225          9)66 

yr.,  at  4^,  is  $9;  then,  $66  will  be  the  in-  .0  4                  7J 

terest  for  as  many  years  as  9  is  contained  $9.0  0 

times  in  66,  which  is  7J,  or  7  yr.  4  mo.  7J  yr.  =  7  yr.  4  mo. 

2.  In  what  time,  at  10%,  will  $500  amount  to  $800? 

OPERATION. 
800 

Solution. -The  interest  will   he   $800  — $500=  ^^^ 

$300.     The  interest  of  $500  for  1  yr.,  at  10^^,  is  $50;  ^^^ 

then,  $300  will  be  the  interest  for  as  many  years  as         10  )  5  00 
50  is  contained  times  in  300,  which  is  6.  ^0 

50)T00 
6 

3.  In  what  time,    at    8%,    will    any    principal    double 
itself? 

Solution.  —  A      principal      has  operation. 

doubled   itself   when    the  interest   be-  8)100 

comes    100  ^.      Since   the   interest   is  1  2^ 

8^   in    1   yr.,   it  will   be   100  f^  in  as 

many   years    as    8    is  contained  times         12^  yr.  =  12  yr.  6  mo. 
in  100,  whicli  is  VJ.\.  ..r  V2  vr.  6  mo. 


SIMPLE  INTEEEST.  233 

Rule. — 1.  Divide  the  given  interest  by  the  interest  oj  the 
prineipal  for  one  year. 

Rem.  1. — If  the  principal  and  amount  are  given,  subtract  the 
principal  from  the  amount  to  find  the  interest, 

Kem.  2. — If  there  be  a  fractional  part  of  a  year  in  the  result,  re- 
duce it  to  months  and  daj^s. 

4.  I  lent  $200,  at  6%,  and  received  $36  interest:  how 
long  was  the  money  lent?  3  yr. 

5.  In  what  time,  at  5%,  will  $60  amount  to  $72? 

4  yr. 

6.  In  what  time,  at  6  % ,  will  any  principal  be  doubled  ? 

16  yr.  8  mo. 

7.  A  man  lent  $375,  at  8%,  and  received  $90  interest: 
how  long  was  it  lent?  3  yr. 

8.  In  what  time,  at  9%,  will  $600  amount  to  $798? 

3  yr.  8  mo. 

9.  In  what  time,  at  10%,  will  any  principal  double 
itself?  10  yr. 

10.  How  long  will  it  take  $250,  at  6%,  to  yield  $34.50 
interest?  2  yr.  3  mo.  18  da. 

11.  How  long  will  it  take  $60,  at  6%,  to  amount  to 
$73.77  ?  3  yr.  9  mo.  27  da. 

12.  How  long  will  it  take  any  principal  to  treble  itself, 
at  6%?  33  yr.  4  mo. 

13^  The  interest  on  $400,  at  7%,  was  $68.60:  how  long 
was  it  loaned?  2  yr.  5  mo.  12  da. 

14.  In  what  time,  at  9%,  will  $700  amount  to  $924.70? 

3  yr.  6  mo.  24  da. 

15.  How  long  will  it  take  any  principal  to  increase 
one-half,  at  8%  ?  6  yr.  3  mo. 

16.  In  whut  time,  at  10^;,  will  $1200  amount  to  $1675? 

•    3  yr.  11  mo.  15  da. 


234  KAY'S  NEW  PKACTICAL  ARITHMETIC. 


CASE   III. 

186.     Given  the    principal,  the    time    and  the  interest, 
to  find  the  per  cent. 

1.  A  merchant  paid  $30  interest  for  the    use    of  S300, 
for  1  yr.  8  mo.:  what  was  the  per  cent? 

OPKRATION. 

Solution.— 1  yr.  8  mo.  are  1|,  or  J  yr.  1  yr.  8  mo.  =  J  yr. 
Since  the  interest  for  f  yr.  is  $30,  the  in-  -Y-  X  |  —  18 

terest  for  1  yr.  is  $18.     $18  is  ^3^  of  $300;  ^y*^  =r:  -^^ 

^^  are  Qfc   (Art.  162).  '    i^jf  =  Qfc' 

2.  At  what   per  cent  will    any  principal    double   itself 
in  20  yr? 

Solution.  —  A   principal    has   doubled    itself  operation. 

when  the  interest  has  become  100 o^.     Since  the      100-^2  0  =  5 
interest  for  20  yr.  is  100  o^,  the  interest  for  1  yr. 

is  ^V  of  100  fc  =  ^  %. 

Bule. — 1.  Find  the  interest  for  one  year,  and  find  what 
per  cent  this  is  of  the  principal. 

3.  I  borrowed  $000  for  2  years  and  paid  $48  interest: 
what  per  cent  did  I  pay?  4^. 

4.  A  broker  paid    $200    interest   for  the  use   of  $1000 
for  2  yr.  6  mo.:  what  was  the  per  cent?  8^. 

5.  The    amount    of  $250    for  2   yr.  4    mo.  24    da.  was 
$310:  what  was  the  per  cent?  10%. 

6.  $23.40    interest  was    paid    for  the  use  of  $260  for  2 
yr. :  what  was  the  per  cent?  4^%- 

7.  At  what    per  cent  will    any  principal    double  itself 
in  12  yr.  6  mo.?  S%. 

8.  The  amount   of  $175  for  3  yr.  7  mo.  was    $250.25: 
what  was  the  per  cent?  12%. 


SIMPLE  INTEREST.  235 

9.  The   interest    of    8450    for    1    yr.    8    mo.    12    da.   is 
$61.20:  what  is  the   per  cent?  8%. 

10.  At  what  per  cent  will  any  principal  double  itself  in 
11  yr.  1  mo.  10  da.?  9%. 

11.  The    amount   of    $650    for    2    yr.    5   mo.  18   da.  is 
$746.20:  what  is  the  per  cent?  6^. 

12.  The  interest  of  $640  for  6  yr.    was  $110.40:  what 
was  the  per  cent?  2^%- 

CASE    IV. 

187.     Given    the    time,  per  cent    and    interest,  to    find 
the  principal. 

1.  The  interest  for   2   3^-.,  at  6%,  is   $27:  what  is  the 
principal  ? 

Solution.— 6  ^0  is -5%  (Art.  162,  3).     Since  operation. 

the  interest  for  1  yr.  is  -5^^  of  the  principal,  for  6  ^  =  -j3_ 

2  yr.  it  is  -^-^  of  the  principal.     Then,  ,^  of  the  J^^  X  f  =  2^5 

principal  are  $27,  2V  ^^  ^^®  principal  is  $9,  and  ^ 

the  principal  is  $225.  ^  X  ^"^  =^  225 

2.  The  interest  for  3  yr.,  at  9%,  is  $21.60:  what  is  the 
principal  ? 

Solution. — ^oL  is  .09.     Since  the  interest  operation. 

for  1  yr.  is  .09,  the  principal,  for  3  yr.  it  is  9  oL  =r.09 

.27,  the  principal.     Then,  the  principal,  mul-  .09  X  3  =.27 

tiplied  by  .27,  is  $21.60,  and  the  principal  is  $21. 60---. 27  =  80 

$21. 60 —  .27  =  $80. 

Rule. — Multiply    the    rate    by   the   time,   and   divide   the 
interest  by  the  product. 

3.  The  interest  for  3  yr.,  at  5%,  is  $8.25:  what  is  the 
principal  ?  $55. 


236  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

4.  The  interest  for  3  yr.,   at  5%,  is    $841.25:  wliat  is 
the  princi2)al?  $2275. 

5.  The   interest  for  1  yr.  4  mo.,  at  6%,  is  $2.20:  what 
is  the  principal?  $28.25. 

6.  What  principal,  at  5%,  will  produce  a  yearly  interest 
of  $1023.75?  $20475. 

7.  What  principal,  at  8%,  will  produce  $30.24  interest 
in  1  yr.  6  mo.  27  da.  ?  $240. 

8.  What  principal,  at  9%,  will  produce  $525.40  interest 
in  12  yr.  3  mo.  20  da.?  $474.40. 

9.  The    interest    for    2    yr.    7    mo.    11    da.,  at   4%,  is 
$9.41:  what  is  the  principal?  $90. 

10.  The   interest   for   5   yr.    8    mo.    24   da.,  at   6%,  is 
$28.38:  what  is  the  principal?  $82.50. 

CASE   V. 

188.     Given    the    time,  per  cent,  and    amount    to    find 
the  principal. 

1.  What   principal    in    5    yr.,  at    G%,  will    amount    to 
$650? 

Solution. — 6^  is  -^jj.    Since  the  interest  for  operation. 

1  yr.  is  3^0  <>f*  the  principal,  for  5  yr.  it  is  j\  of  (j  cL  =  ^-^ 

the   principal,   and   the   amount   is  \^   of   the  ,3yXf  =  A 

principal.     Then,  }|  of  the  principal  are  $650,  }g -f  fV^if 

y^^  of  the  principal  is  $50,  and  the  principal  is  50 

$500.  IMx^$='jOO 

Rule. — Multiply   the    rate    by    the    time,  and    divide   the 
amount  by  1  -\-  the  j)roduct. 

2.  What   principal    in    9    yr.,  at    5%,   will    amount   to 
$435?  $300. 


COMPOUND  INTP]REST.  237 

3.  The  amount    for  4  yr.,  at   5%,  is  $571.20:    what  is 
the  interest?  S95.20. 

4.  The  amount    for  6  yr.,  at  7%,  is  $532.50:    what  is 
the  interest?  $157.50. 

5.  The    amount    for    2    yr.  9    mo.,  at  8%,  is    $285.48: 
what  is  the  j)rincipal?  $234. 

6.  The  amount  for  2  yr.  6  mo.,  at  6%,  is  $690:  what 
is  the  interest?  $90.  ' 

7.  The   amount    for    3    yr.    4    mo.    24    da.,    at    7%,    is 
$643.76:  what  is  the  principal?  $520. 

8.  The    amount    for    4    yr.    3    mo.    27    da.,   at    4%,   is 
$914.94:  what  is  the  interest?  $134.94. 

189,  Formulas  for  the  five  cases  of  Interest. 

Let  h  represent  the  principal,  t  the  time,  r  the  rate,  and 
i  the  interest.     Then, 

Case      I.  b  Xr  Xt=^i. 

Case    II.  I  -f-  (6  X  0  —  ^• 

Case  TIT.  (i  ^  t)  ~  b  =  r. 

Case  lY.  i  -^{rXt)  =  b,  *       ' 

Case    Y.     ^--^"^-^^^b. 

1  +  (^-  X  0 


COMPOUND   INTEREST. 

190.  In  Compound  Interest  the  principal  is  increased 
yearly  b}^  the  addition  of  the  interest. 

Rem.  1.  —  Sometimes  tlie  interest  is  added  semi-amiually,  or 
quarterly. 

Rem.  2. — The  way  in  which  interest  is  legally  compounded  is,  at 
the  end  of  each  year,  to  take  up  the  old  note  and  give  a  new  one 
with  a  face  equal  to  both  the  principal  and  interest  of  the  former 
note. 


238  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

1.   Find  the    compound  interest    of  $300    for  3   3^r.,  at 
6%. 

Solution. — The  interest 
of  $300  for  1  yr.,  at  6^,  is  operation. 

$18;    the  amount  is  $18+  $3  0  0  $3  3  7.0  8 

$300  r^  $318.    The   interest  0  6  .0  6 

of  $318   for    1    yr.,  at  6^^,  18.0  0  2  0.2  2  4  8 

is    $19.08;    th'     amount    is  3  00  3  3  7.0  8 

$10.08  +  $318  =  $337.08.  -3X8.  357^3048 

The  interest  of  $337.08  for  .0  6  3  0  0 


18.0  0 

300 
3  18. 

.0  6 

1  9.0  8 

318 

8  3  7.08 

1   yr.,  at    6^^,   is  $20.2248;  19.08  $57.3048 

the  amount  is  $20.2248  + 
$337.08  =  $357.3048.  Then, 
the  compound  interest  is 
$357.3048  —  $300  =  $57.30. 


Kule. — 1.  Find  the  amount  of  the  given  principal  for 
the  first  yearj  and  make  it  the  principal  for  the  second 
year, 

2.  Find  the  amount  of  this  principal  for  the  second  year^ 
make  it  the  principal  for  the  third  year,  and  so  on  for  the 
given  number  of  yearns. 

3.  From  the  last  amount  subtract  the  given  principal ;  the 
remainder  will   be   the  compound  interest. 

Rem.  1. — When  the  interest  is  payable  half-yearly,  or  quarterly, 
find  the  interest  for  a  half,  or  a  quarter  year,  and  proceed  in  other 
respects  as  when  the  interest  is  payable  3'^early. 

Rem.  2. — When  the  time  is  years,  months,  and  days,  find  the 
amount  for  the  years,  then  compute  the  interest  on  this  for  the 
months  and  days,  and  add  it  to  the  last  amount. 

Find  the  amount,  at  6%,  compound  interest, 

2.  Of  $500,  for  3  years.  S595.51. 

3.  Of  $800,  for  4  3^ears.  $1009.98. 


ANNUAL  INTEREST.  239 

Find  the  compound  interest 

4.  Of  $250,  for  3  yr.,  at  6%.  $47.75. 

5.  Of  $300,  for  4  yr.,  at  5%.  $64.65. 

6.  Of  $200,  for  2  yr.,  at  6%,  pajable  semi-annually. 

$25.10. 

7.  Find  the  amount  of  $500,  for  2  yr.,  at  20%  com- 
pound interest,  payable  quarterly.  $738.73. 

8.  What  is  the  compound  interest  of  $300,  for  2  yr.  6 
mo.,  at  6%?  $47.19. 

9.  What  is  the  compound  interest  of  $1000,  for  2  yr. 
8  mo.  15  da.,  at  6%?  $171.35. 

10.  What  is  the  amount  of  $620  at  compound  interest 
semi-annually,  for  3  yr.  6  mo.,  at  6%?  $762.52. 

11.  What  is  the  diiference  between  simple  interest  and 
compound  interest  on  $500,  for  4  yr.  8  mo.,  at  6%  ? 

$16.49. 

ANNUAL  INTEREST. 

191.  Annual  Interest  is  interest  on  the  principal, 
and  on  each  annual  interest  after  it  is  due. 

Rem.  1. — This  interest  is  sometimes  semi-annual  or  quarterly. 

Rem.  2. — Annual  interest  may  be  collected  when  the  note  or  bond, 
reads  "  with  interest  payable  annually." 

Rem.  3. — The  annual  interest  is  sometimes  represented  by  interest 
notes;  these  are  given  at  the  same  time  as  the  note  for  the  principal, 
and  draw  interest  if  not  paid  when  due. 

Rem.  4. — The  annual  interest  on  bonds  is  sometimes  represented  by 
interest  notes,  called  coujwns;  these  are  detached  from  the  bond  and 
presented  for  payment  when  the  interest  is  due. 

1.  ]^o  interest  having  been  paid,  find  the  amount  due 
in  4  yr.  8  mo.  24  da.,  on  a  note  for  $400,  with  interest 
at  6%,  payable   annually. 


OPERATION. 

2  ):5  6  8 

400 

.2  8  4 

.0  6 

400 

2  4.00 

1  1  3.6  0  0 

2)1.072 

yr.  mo.  da. 

.5  3  6 

3     8     24 

24 

2     8     24 

2144 

1     8     24 

1072 

8    24 

1  !\8  6  4 

8  11      0 

1  1  3.6  0 

400. 

$5  2  6.4  6  4 

240  KAY'S  NEW  PKACTICAL  ARITHMETIC. 


Solution. — The  interest  of 
$400  for  4  yr.  8  mo.  24  da.,  at 
6^,  is  $113.60.  One  annual  in- 
terest of  $400,  at  6^c»  is  $24. 
The  first  annual  interest  re- 
mains  unpaid  3  yr.  8  mo.  24 
da.;  the  second,  2  yr.  8  mo. 
24  da.;  the  third,  1  yr.  8  mo. 
24  da.,  and  the  fourth,  8  mo. 
24  da.;  hence,  interest  must  be 
reckoned  on  $24  for  8  yr.  11 
mo.  6  da.;  this  is  $12,864. 
The  amount,  then,  is  $12,864 
-f  $113.60  -I-  $100  =  $526.46. 


Bule. — 1.  Find  the  interest  of  the  principal  for  the  time 
during  xchich  no  annual  interest  is  paid. 

2.  Find  the  interest  of  one  annual  interest  for  the  sum  of 
the  times  each  annual  interest  remains  unpaid. 

3.  The  sum  of  the  two  interests  will  be  the  interest  due, 
and  this,  added  to  the  principal,  will  be  the  amount  due. 


2.  ^o  interest  having  been  paid,  find  the  amount  due 
in  3  yr.  on  a  note  for  S800,  with  Interest  at  8%,  payable 
annually.  S1007.36. 

3.  The  interest  having  been  paid  for  2  yr.,  find  the 
amount  due  in  5  yr.  on  a  note  for  $750,  with  interest 
at  10%,  payable  annually.  8997.50. 

4.  No  interest  having  been  paid  for  4  yr.,  find  the 
interest  due  in  6  yr.  on  a  bond  for  $10000,  with  interest 
at  5%,  payable  annually.  $2150. 

5.  No  interest  having  been  paid,  find  the  amount  due 
Sept.  1,  1877,  on  a  note  for  $500,  dated  June  1,  1875,  with 
interest  at  6%,  payable  semi-annually.  $571.10. 


PARTIAL  PAYMENTS.  241 

6.  [S1200.]  Milwaukee,  Wis.,  May  12,  1873. 
For  value  received,  on  demand,  I  promise  to  pay  John 

G.  Morgan,  or  order,  twelve  hundred  dollars,  with  interest 
at  6%,  payable  annually.  H.  W.  Slocum. 

No  interest  having  been  paid,  what  was  the  amount 
due  on  this  note,  Sei:)tember  20,  1877?  $1545.66. 

7.  [$1500.]         New  Orleans,  La.,  October  10,  1872. 
On   the   first    day  of  May,   1877,  for  value  received,  I 

promise  to  pay  Andrew^  Jackson,   or  order,  fifteen   hun- 
dred dollars,  with  interest,  payable  annually,  at  5%. 

George  Quitman. 
No  interest  having  been   paid,  what   amount  was  due 
at  maturity?  $1872.75. 

8.  What  is  the  difierence  between  simple  and  annual 
interest  on  $1000  for  5  yr.,  at  6%  ?  $36. 

9.  What  will  be  due  on  six  500-dollar  city  bonds  run- 
ning 3  yr.,  with  interest  at  6%,  payable  semi-annually, 
if  the  interest  should  not  be  paid?  $3580.50. 

10.  The  interest  on  U.  S.  4  per  cent  bonds  is  payable 
quarterly  in  gold ;  granting  that  the  income  from  them 
might  be  immediately  invested,  at  6%,  w^hat  would  the 
income  on  20  1000-dollar  bonds  amount  to  in  5  yr., 
with  gold  at  105?  $4798.50. 


PARTIAL   PAYMENTS. 

192.  1.  A  partial  payment  is  a  sum  of  money,  less 
than  the  face,  paid  on  a  note. 

2.  The  receipt  of  a  partial  payment  is  acknowledged 
by  indorsing  it  on  the  back  of  the  note. 

Prac.  16. 


242  KAY'S  NEW  TKACTICAL  ARITHMETIC. 

3.  The  indorsement  consists  of  the  date  and  amount 
of  the  pajMuent. 

4.  The  rule  of  the  Supreme  Court  of  the  United  States, 
in  reference  to  Partial  Payments,  is  as  follows : 

'^  When  partial  payments  have  been  made,  apply  the 
payment,  in  the  first  place,  to  the  discharge  of  the  interest 
then  due. 

^'  If  the  payment  exceeds  the  interest,  the  surplus  goes 
toward  discharging  the  principal,  and  the  subsequent  in- 
terest is  to  be  computed  on  the  balance  of  principal  remuin- 
ing  due. 

"  If  the  payment  be  less  than  the  interest,  the  surplus  of 
interest  must  not  be  taken  to  augment  the  principal,  but 
interest  continues  on  the  former  principal,  tintil  the  period 
when  the  payments,  taken  together,  exceed  the  interest  due, 
and  then  the  surplus  is  to  be  applied  toward  discharging 
the  principal ;  and  interest  is  to  be  computed  on  the  balance, 
as  aforesaid.'' — Kent,  C.  J. 

Rem. — This  rule  is  founded  on  the  principle  that  neither  interest 
nor  payment  shall  draw  interest. 

1.  [SIOOO.]  Boston,  Mass.,  May  1,  1875. 

For  value  received,  on  demand,  I  promise  to  pa}^  to 
Alonzo  Warren,  or  order,  one  thousand  dollars,  with 
interest  at  6%.  William  Murdock. 

On  this  note  partial  payments  were  indorsed  as  fol- 
lows : 

November  25,  1875,  $134;  March  7,  1876,  $315.30; 
August  13,  1876,  $15.60;  June  1,  1877,  $25;  April  25, 
1878,  $236.20.  What  w^as  the  amount  due  on  settlement, 
September  10,  1878? 


FAUTIAL  PAYMENTS. 


243 


Solution.  —The  time 
from  May  1,  1875,  to 
November  25,  1875,  is  6 
mo.  24  da.;  the  interest 
of  $1000  for  this  time  is 
$34;  the  payment,  $134, 
exceeds  the  interest;  the 
amount  is  $1034;  $1034 
—  $134  =  $900,  the  sec- 
ond  iirincijiol. 

The  time  from  No- 
vember 25,  1875,  to 
March  7,  1876,  is  3  mo. 
12  da.;  the  interest  of 
$900  for  this  time  is 
$15.30;  the  payment, 
$315.30,  exceeds  the  in- 
terest; the  amount  is 
$915.30;  $915.30—1315.30 
=r$600,  the  third  jirin- 
cipal. 

The  time  from  March 
7,  1876,  to  August  13, 
1876,  is  5  mo.  6  da.;  the 
interest  of  $600  for  this 
time  is  $15.60;  the  pay- 
ment, $15.60,  equals  the 
interest;  the  amount  is 
$615.60;  $615.60— $15.60 
^^  $600,  the  fourth  prin- 
cipal. 

The  time  from  August 
13,  1876,  to  June  1,  1877, 
is  9  mo.  18  da.;  the  in- 
terest of  $600  for  this 
time  is  $28.80;  the  pay- 
ment, $25,  is  less  than 
the  interest;  continue  to 
princij^al. 


OPERATION. 

1875 

11   25 

$134 

1000 

1875 

5     1 

.0  3  4 

6  24 

34 

2  ).0  6  8 
.0  3  4 

1000 

1034 

134 

1876 

3   7 

$  3  1  5.3  0 

9  00 

1875 

1  1  2  5 

.017 

3  12 

15.3  0 

2  ).0  3  4 
.017 

900 
9  1  5.3  0 
3  1  5.3  0 

1876 
187  6 

8  13 
3   7 
5   6 

$15.6  0 

600 
.0  2  6 

15.6  0 
600 

2).0  5  2 

6  1  5.6  0 

.02  6 

15.6  0 

1877 

6   1 

$25 

600 

.0  4  8 

.2  8.8  0 

1876 

8  13 

9  18 

2).0  9  6 

.0  4  8 

600 

1878 
1877 

4  2  5 

6   1 

10  2  4 

$  2  3  6.2  0 
2  6  1.2  0 

.0  5  4 
3  2.4  0 
2  8.8  0 
6  1.2  0 

2).108 

6  00. 

.0  5  4 

6  6  1.2  0 
2  6  1.2  0 

1878 

9  10 

400 

1878 

4  25 

.0225 

4  15 

9 

2).0  4  5 
.0225 

400 

409 

find    the   interest    on 


the  fourth 


244  KAY'S  NEW  PRACTICxVL  AKITIIMETIC. 

The  time  from  June  1,  1877,  to  April  25,  1878,  is  10  mo.  24  da.; 
the  interest  of  $000  for  this  time  is  $32.40;  the  sum  of  the  payments, 
$261.20,  exceeds  the  sum  of  the  interests,  $01.20;  the  amount  is 
$661.20;  $661.20  — $261. 20  ==$400,  the  ffth  principal. 

The  time  from  April  25,  1878,  to  September  10,  1878,  is  4  mo.  15 
<;la.;  the  interest  of  $400  for  this  time  is  $9;  the  amount  due  on  set- 
tlement is  $409. 

RULE. 

I.  When  each  payment  equals  or  exceeds  the  interest. 

1.  Find  the  time  from  the  date  of  the  note  to  the  date 
of  the  first  payment 

2.  Find  the  amount  of  the  given  j^rincipal  for  this  time. 

3.  From  this  amount  subtract  the  payment ;  the  remainder 
is  the  second  principal. 

4.  Find  the  time  from  the  date  of  the  first  payment  to 
the  date  of  the  second  payment. 

5.  Then  proceed  with  the  second  principal  as  with  the 
first,  anil  so  on  to  the  date  of  settlement. 

II.  When  one  or  more  payments  are  less  than  the 
interest. 

1.  Continue  to  find  the  interest  on  the  same  principal 
until  a  date  is  reached,  when  the  sum  of  the  payments 
equals  or  exceeds  the  sum  of  the  interests. 

2.  Then  subtract  the  sum  of  the  payments  from  the 
amount;  the  remainder  is  th?.  next  principal. 

Rem.— Sometimes  it  may  be  determined,  by  inspection,  that  the 
payment  is  less  than  the  interest;  when  this  can  be  done,  it  is  not 
necessary  to  find  the  intermediate  time  and  interest,  but  interest  may 
at  once  be  found  to  the  date  when  it  is  apparent  that  the  sum  of  the 
payments  exceeds  the  interest. 


PARTIAL  PAYMENTS.  245 

2.  [S350.]  Boston,  Mass.,  July  1,  1875. 
For  value  received,  I  promise  to  pay  Edward  Sargent, 

or    order,   on    demand,  three    hundred    and    fifty  doHars, 
with  interest  at  6%.  James  Gordon. 

Indorsements:  March  1,  1876,  ^44;  October  1,  1876, 
SIO  ;  January  1,  1877,  S26  ;  December  1,  1877,  $15.  What 
w^as  the  amount  due  on  settlement,  March  16,  1878? 

$306.75. 

3.  A  note  of  $200  is  dated  January  1,  1875.  Indorse- 
ment: January  1,  1876,  $70.  What  was  the  amount  due 
January  1,  1877,  interest  at  6%  ?  $150.52. 

4.  A  note  of  $300  is  dated  July  1,  1873.  Indorsements: 
January  1,  1874,  $109;  July  1,  1874,  $100.  What  was 
the  amount  due  January  1,  1875,  interest  at  6%? 

$109.18. 

5.  A  note  of  $150  is  dated  May  10,  1870.  Indorse- 
ments: September  10,  1871,  $32;  September  10,  1872, 
$6.80.  What  was  the  amount  due  November  10,  1872, 
interest  at  6%  ?  $132.30. 

6.  A  note  of  $200  is  dated  March  5,  1871.  Indorse- 
ments: June  5,  1872,  $20;  December  5,  1872,  $50.50. 
What  was  the  amount  due  June  5,  1874,  interest  at  10%  ? 

$189.18. 

7.  A  note  of  $250  is  dated  January  1,  1875.  Indorse- 
ments: June  1,  1875,  $6;  January  1,  1876,  $21.50 
What  was  the  amount  due  July  1,  1876,  interest  at  7%' 

$248.40, 

8.  A  note  of  $180  is  dated  August  1,  1874.  Indorse 
ments:  February  1,  1875,  $25.40;  August  1,  1875,  $4.30 
January  1,  1876,  $30.  What  was  the  amount  due  July 
1,  1876,  interest  at  6%  ?  $138.54. 

9.  A  note    of  $400    is  dated  March  1,  1875.     Indorse 
ments:     September  1,  1875,  $10;    January  1,  1876,   $30 


24G  RAY'S  NEW  PRACTICAL  ARITHMETIC. 


July  1,    1876,  $11;    September    1,  1876,  $80.     What  was 
the  amount  due  March  1,  1877,.  iuterest  at  6%? 

8313.33. 

10.  A  note  of  $450  is  dated  April  16,  1876.  Indorse- 
ments: January  1,  1877,  $20;  April  1,  1877,  $14;  July 
16,  1877,  $31;  December  25;  1877,  $10;  July4,  1878, 
$18.  What  was  the  amount  due  June  1,  1879,  interest 
at  8%  ?  $466.50. 

11.  A  note  of  $1000  is  dated  January  1,  1870.  In- 
dorsements: May  1,  1870,  $18;  September  4,  1870,  $20; 
December  16,  1870,  $15;  April  10,  1871,  $21;  July  13, 
1871,  $118;  December  23,  1871,  $324.  What  was  the 
amount  due  October  1,  1873,  interest  at  6^?       $663.80. 

193.  When  partial  paj'ments  are  made  on  notes  and 
accounts  running  a  3'ear  or  less,  the  amount  due  is 
commonly  found  by  the 

MERCANTILE   RULE. 

1.  Find  the  amount  of  the  principal  from  the  date  of  the 
note  to  the  date  of  settlement. 

2.  Find  the  amount  of  each  payment  from  its  date  to  the 
date  of  settlement. 

3.  From  the  amount  of  the  principal  subtract  the  sum  of 
the  amounts  of  the  payments. 

1.  A  note  of  $320  is  dated  Jan.  1,  1876.  Indorsements: 
May  1,  1876,  $50;  Nov.  16,  1876,  $100.  What  was  the 
amount  due  Jan.  1,  1877,  interest  at  6^  ?  $186.45. 

2.  An  account  of  $540  was  due  March  1,  1877. 
Credits:  May  1,  1877,  $90;  July  1,  1877,  $100;  Aug.  1, 
1877,  $150;  Oct.  11,  1877,  $180.  W^hat  was  the  amount 
due  on  settlement  Jan.  1,  1878,  interest  at  S%?         $39. 


DEFINITIONS. 

194.  1.  Discount  is  interest  paid  in  advance. 

2.  There  are  two  kinds  of  discount,  Bank  Discount  and 
True  Discount. 

BANK   DISCOUNT. 

195.  1.  Banks  lend  money  on  two  sorts  of  notes.  (1) 
accommodation  notes,  and   (2)  business  notes. 

Rem. — These  notes  are  frequently  termed  accommodation  paper, 
and  business  paper. 

2.  An  accommodation  note  is  made  payable  to  the 
bank  which  lends  the  money. 

TvEM. — The  following  is  a  common  form  of  an  accommodation  note: 

^500.  Chicago,  III.,  October  20,  1877. 

Ninety  days  after  date,  we,  or  either  of  us,  promise  to 
pay  to  the  Second  National  Bank  of  Chicago,  111.,  ^ve 
hundred   dollars,   for   value   received. 

O.  S.  West. 

W.  B.  Sharp. 

18  / 
Due  January       /oi  ,  1878. 

•^         ^^^  (247) 


248  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

3.  A  business  note  is  payable  to  an  individual. 

4.  A  business  note  may  be  nefjotiahle  or  not  negotiable. 

5.  A  negotiable  note  is  one  that  can  be  bought  and 
sold. 

Rem. — The  followinu:;  aro  common  forms  of  business  notes: 

1st.    JV'ot  ne^otiahlc. 

$200.  Buffalo,  N.  Y.,  March  21,  1877. 

On  demand,  I  promise  to  pay  Charles  II.  Peek,  two 
hundred  dollars,  ibr  value  received. 

G.   W.  Clinton. 

This  note  is  payahle  only  to  Charles  H.  Peck;  it  is  due  at  once, 
and  bears  interest  from  datt?. 

2d.   JS^egotidble, 

$1000.  St.  Louis,  Mo.,  May  1,  1877. 

One  year  after  date,  I  promise  to  pay  to  David  King, 
or  order,  one  thousand  dollars,  for  value  received. 

Elmer  B.  Archer. 

The  words  "or  order"  make  this  note  negotiable.  If  David  King 
transfers  it,  he  must  indorse  it — that  is,  write  his  name  across  the  back 
of  it.     This  note  bears  no  interest  till  after  it  is  due. 

3d.  JYeiotiable. 

$150.  Washington,  D.  C,  August  10,  1877. 

On  or  before  the  first  day  of  May,  1878.  I  promise  to 
pay  Amos  Durand,  or  bearer,  one  hundred  and  fifty 
dollars,  with  interest  at  10^  from  date,  for  value  received. 

John  Sherwood. 

The  words  "  or  bearer  "  make  this  note  negotiable  without  indorse- 
ment.    This  note  bears  interest  from  date,  it  being  so  specified. 


PANK  DISCOUNT.  249 

A 

6.  A  note  is  payable,  or  nominally  due,  at  the  end  of 
the  time  specified  in  the  note. 

7.  A  note  matures,  or  is  legally  due,  three  days  after 
the  specified  time. 

8.  The  three  days  after  the  specified  time  are  called 
days  of  grace. 

Rem.  1. — Banks  lend  money  only  on  short  time;  rarely  beyond  3 
months. 

To  find  ivhen  a  note  matures: 

1st.  When  the  time  is  expressed  in  days : 

Rule. —  Count  the  days  from  the  date  of  the  note  and  add 
three  days. 

2d.  When  the  time  is  expressed  in  months: 

Rule. —  Count  the  months  from  the  date  and  add  three 
days. 

Rem.  2. — In  Delaware,  Maryland,  Pennsylvania,  Missouri,  and 
the  District  of  Columbia,  the  day  of  discou7it  is  the  first  day  of  the 
time. 

Rem.  8. — When  a  note  in  bank  is  not  paid  at  maturity,  it  goes  to 
protest — that  is,  a  written  notice  of  this  fact,  made  out  in  legal  form, 
by  a  notary  public,  is  served  on  the  indorsers,  or  securit^^ 

9.  The  bank  discount  is  simple  interest  taken  in 
advance. 

10.  The  proceeds  is  the    money  received  on  the  note. 

11.  In  Bank  Discount  four  quantities  are  considered: 
(1)  The /ace  of  the  note,  (2)  the  per  cent,  (3)  the  time^ 
and  (4)  the  discount. 

12.  Any  three  of  these  quantities  being  given,  the 
fourth  may  be    found.     We  will    consider  two   cases. 


250  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

CASE    I. 

196.  Given  the  face  of  the  note,  the  per  cent,  and 
the  time  to  find  the  discount  and  the  proceeds. 

1st.    When  the  note  does  not  hear  interest. 

1.  Find  the  date  when  due,  bank  discount,  and  pro- 
ceeds of  the  following  accommodation  note,  discounted 
at   6%: 

$700.  Mobile,  Ala.,  June  25,  1877. 

Sixty  days  after  date  we,  or  either  of  us,  promise  to 
pay  to  the  First  National  Bank,  of  Mobile,  Ala.,  seven 
hundred  dollars  for  value  received. 

Charles  Walker. 
Walter   Smith. 

OPERATION. 

Solution. — The  note  is  due   August      /oy  ,  nTTTF 

/  ^ '  .0  1  0  o 

1877  (Art.  78).     The  interest  of  $1  for  63  days,  7  00 

at   6^,  is  $0.0105,  and  the  interest   of  $700   is  73  5  0  0 

$0.0105  X  700  =  $7.35;  this  is  the  discount ;  then,  7  0  0  0  0~ 

$700  —  $7.35  =  $692.65,  the  proceeds.  7*3  ^ 

ir92ir5 

Rule. — 1.  Find  the  interest  on  the  face  of  the  note  for 
the  given  time;  this  is  the  hank  discount. 

2.  From  the  face  of  the  note  subtract  the  discount ;  the 
remainder  is  the  proceeds. 

Find  the  date  when  due,  bank  discount,  and  pro- 
ceeds   of 

2.  A  note  of  $100,  dated  June  20,  payable  in  60  days, 
and  discounted  at  6%.  August  ^  /22'  SI. 05,  $98.95. 


BANK  DISCOUNT.  251 

3.  A  note  of  $120,   dated   October   12,  payable    in    30 
(lays,  and  discounted  at  8^. 

November  ^7^4,  S0.88,  $119.12. 

4.  A   note   of  $140,   dated   January    15,   payable   in   4 
months,  and  discounted   at  6^. 

May  ^^jig,  $2.87,  $137.13. 

5.  A    note    of    $180,    dated    April    10,    payable    in    6 
months,  and  discounted  at  4%. 

October   ^^/^3,  $3.66,  $176.34. 

6.  A  note    of  $250,  dated    December    1,  payable    in    5 
months,  and  discounted  at  8^. 

May  y^,  $8.50,  $241.50. 

7.  A   note    of  $375,   dated    August   4,    payable    in    30 
days,  and  discounted  at  6%. 

September  ^g  ,  $2.06,  $372.94. 

8.  A  note    of  $600,  dated    February  12,  1876,  paj^able 
in  60  days,  and  discounted  at  9%. 

April  ^^^^5  ,  $9.45,  $590.55. 

9.  A  note  of  $1200,  dated  February  20,  1877,  payable 
in  90  days,  and  discounted  at  10^. 

May  ^^24'  ^^1'  ^^^^^• 

10.  A  note  of  $1780,  dated   January  11,  1872,  payable 
in  90  days,  and  discounted  at  6^. 

April  ^^/i3,  $27.59,  $1752.41. 


Find  the  date  when    due,  time  of  discount,  bank  dis- 
count, and  proceeds  of  the  following  business  notes: 


252  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

11.  [$600.]  San  Francisco,  Cal.,  Sept.  15,  1876. 
One  3^ear  after  date,  I  promise  to  pa}^  to  the  order  of 

Abel    E.  Worth,    at    the    First    National    Bank    of   San 
Francisco,  Cal.,  six  hundred  dollars,  for  value  received. 

George  M.  Burgess. 

Discounted  May  21,  1877,  at  10%. 

Sept.  ^^/-^g,  1877,  120  days,  S20,  $580. 

12.  [SIOOO.]  Nashville,  Tenn.,  May  8,  1877. 
Ninety   days   after   date,    I  promise   to  pay  Albert    E. 

Kirk,  or  order,  one  thousand  dollars,  for  value  received. 

Jacob  Simmons. 
Discounted  June  22,  1877,  at  6%. 

August    /g,  48  days,  $8,  $992. 

13.  [S1500.]  Pittsburgh,  Pa.,  July  10,  1877. 
Six    months    after    date,    I    promise    to    pa}''   Alex.    M. 

Guthrie,    or    bearer,    fifteen    hundred    dollars,    for    value 
received.  Orlando  Watson. 

Discounted  October  25,  1877,  at  6^. 

January  ^^3,  1878,  81  days,  $20.25,  $1479.75. 

2d.    When  the  note  hears  interest. 

1.  Find  the  dale  when  due,  time  of  discount,  bank 
discount,  and  proceeds  of  the  following  business  note: 

$800.  Dayton,  O.,  January  5,  1877. 

Six  months  after  date,  I  promise  to  pay  to  the  order 
of  Charles  Stuart,  at  the  Dayton  National  Bank,  of  Day- 
ton, O.,  eight  hundred  dollars,  with  interest  at  6^,  for 
value  received.  Francis  Murphy. 


BANK  DISCOUNT.  253 

Discounted  April  15,  1877,  at  8^. 


OPERATION. 

2).0  6  1 

800 

.0  30  5 

.0305 

2  4.4  00  0 

3)84 

800 

3).0  2  8 

824T0 

.0  0  9^ 

.01  8f 

2 

5  4  9  6  0 

.0181 

659520 

82440 

82  4.4  0 

1  5.3  8  8  8  0 

1  5.3  9 

$8  0  9.0  1 

Solution. — The  note  is  due 
July  ^/g ,  1877.  The  time  of  dis- 
count, from  April  15  to  July  8, 
is  84  days.  The  amount  of  $800 
for  6  mo.  3  da.,  at  ^o]^,  is  $824.40. 
The  bank  discount  of  $824.40  for 
84  days,  at  '^c/^,  is  $15.39.  The 
j^roceeds  are  $809.01. 


Rule. — 1.  Find  the  amount  of  the  note  for  the  given  time. 
2.  Find  the  bank  discount  and  proceeds  of  this  amount. 

Eem.  —  In  the  following  examples,  remember  that  in  leap  years 
February  has  29  days. 

Find  the  date  when  due,  time  of  discount,  bank  dis- 
count, and  proceeds  of 

2.  A  note  of  $150,  dated  May  20,  1875,  payable  in  6 
months,  with  interest    at  6%,  and  discounted  September 

9,  1875,  at  S%. 

November  ^^/23 ,  1875,  75  days,  $2.58,  $152. 

3.  A  note  of  $300,  dated  August  5,  1876,  payable  in 
1  year,  with    interest    at    S%,  and    discounted   April  16, 

1877,  at  6%. 

August  ^g,  1877,  114  days,  $6.16,  $318.04. 

4.  A  note  of  $450,  dated  March  4,  1877,  due  January 
1,  1878,  with  interest  at  6%,  and  discounted  August  13, 
1877,  at  10%. 

January  ^^4,  1878,  144  days,  $18.90,  $453.60. 


254  KAY'S  NEW  Pll ACTIO AL  ARITHMETIC. 

5.  A  note  of  $650,  dated  May  16,  1876,  due  Sept.  1, 
1878,  with  interest  at  9%,  and  discounted  April  25,  1878, 
at  6%. 

Sept.  Y4,   1878,  132  days,  $17.26.     $767.29. 


6.  A  note  of  $840,  dated  September  1,  1875,  payable 
in  6  months,  with  interest  at  10^,  and  discounted  De- 
cember 20,  1875,  at  8%. 

March  Y4,  1876,  75  days,  $14.71,  $867.99. 

7.  A  note  of  $1400,  dated  July  19,  1875,  due  May  1, 
1876,  with  interest  at  6^,  and  discounted  Jan.  17,  1876, 
at  10%. 

May  y4,  1876,  108  days,  $44,  $1422.50. 

8.  A  note  of  $2400,  dated  Oct.  16,  1876,  due  Jan.  1, 
1878,  with  interest  at  8^,  and  discounted  July  26,  1877, 
at  10%. 

January  '^/ji^,  1878,  162  days,  $118.51,  $2515.09. 

9.  [$3500.]  Macon,  Ala.,  October,  15,  1877. 
One   year   after    date,  I  promise  to  pay  Adam  Moore, 

or    order,  thirty-five    hundred    dollars,    with    interest    at 
6%,  for  value  received.  Joseph  Stephens. 

Discounted  May  15,  1878,  at  9^. 

October  ^^j^g,  1878,   156   days,  $144.76,  $3566.99. 

10.  [$6000.]  Frankfort,  Ky.,  3Iay  10,  1875. 
One  year  after  date,  1  promise  to  pay  Henry  Warren, 

or  order,  six  thousand  dollars,  with   interest   at   S%.  for 
value  received.  Amos  E.  Burton. 

Discounted  November  21,  1875,  at  10^. 

May  ^^L,  1876,  174  days,  $313.39,  $6170.61. 


BANK  DISCOUNT.  255 


CASE  II. 

197e  Given  the  per  cent,  the  time,  and  the  proceeds, 
to  find  the  face  of  the  note. 

1.  For  what  sum  due  90  days  hence,  must  I  give  a 
note  to  a  bank,  that,  when  discounted  at  6%,  the  pra 
ceeds  will  be  S177.21? 

OPERATION. 

3)93  1.0  00  0 

Solution.— The  bank  discount  ^  )>0  3 1  -0155 

of  $1  for  93  days,  at  6^^,  is  $0.0155,  .0155  .9845 

and  the  proceeds  $1  — $0.0155  — 

$0.9845.    Then,  $177.21  is  the  pro-         .0845)177.21(180 
ceeds  of  177.21  --  .9845  =  $180.  9845 

787CO 
78760 


Rule.— 1.  Find  the  proceeds  of  %l  for  the  given  time  at 
the  given  per  cent. 

2.  By  this  divide  the  given  proceeds. 

2.  The  proceeds  of  a  note  discounted  at  a  bank  for  GO 
days,  at  6^,  were  $197.90:  what  was  the  face  of  the 
note?  $200. 

3.  For  what  sum  must  a  note  be  made,  so  that  when 
discounted  at  a  bank,  for  90  days,  at  6%,  the  proceeds 
will  be  $393.80?  $400. 

4.  What  must  be  the  face  of  a  note,  that  when  dis- 
counted at  a  bank  for  5  months^  at  8^,  the  proceeds 
may  be  $217.35?  $225. 

5.  The  proceeds  of  a  note  are  $352.62,  the  time  4 
months,  and  the  discount  at  (3^  :  what  is  the  face? 

S360. 


256  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

6.  1  wish  to  borrow  $400  from  a  bank  for  30  days: 
what  must  be  the  face  of  my  note,  that,  when  discounted 
at  6%,  I  may  receive  this  amount?  $402.21. 

7.  I  wish  to  obtain  from  a  bank  $500  for  60  days: 
for  what  sum  must  I  give  my  note,  at  8^  discount? 

$507.10. 

8.  I  wish  to  use  $1500  for  6  months;  if  I  can  obtain 
money  from  a  bank,  at  a  discount  of  10%,  for  what 
sum  must  I  give  my  note  to  realize  this  amount? 

$1580.33. 

9.  A  note  dated  February  19,  1876,  payable  January 
1,  1877,  and  bearing  8%  interest,  was  discounted  Octo- 
ber 12,  1876,  at  6fc  ;  the  proceeds  were  $1055.02:  what 
was  the  face  of  the  note?  $1000. 


TRUE  DISCOUNT. 

198.  1.  The  present  worth  of  a  note  is  a  sum  of 
money,  which,  being  on  interest  for  the  given  time  at  a 
given  i)er  cent,  will  amount  to  the  same  as  the  note. 

2.  The  true  discount  is  the  difference  between  the 
present  worth  and  the  amount  of  the  note. 

Rem.  1. — Notes,  d^^bts,  and  running  accounts  are  discounted  by 
True  Discount. 

Rem.  2. — Banks  sometimes  discount  by  the  method  of  True 
Discount. 

199.  Given  the  face  of  the  note,  the  time,  and  the  per 
cent,  to  find  the  present  worth  and  discount. 

1.  Find  the  present  worth  and  discount,  at  6^,  of  a 
note  of  $430.50,  due  in  2  yr.  5  mo.  18  da. 


TRUE  DISCOUNT. 


257 


Solution. — The  amount  of  $1  for 
2  yr.  5  mo.  18  da.,  at  6^o,  is  $1,148. 
Then,  the  present  worth  of  $430.50 
is  430.50 --1.148  =  $375;  and  the  dis- 
count is  $430.50  — $375  =  $55.50. 


OPEKATIOK. 

2  ).2  9  6 

.14  8 

1.0  0 

1.14  8 

.1  4  8  )  4  3  0.5  0  (  3  7  5 

3444 

8610 

8036 

5740 

5  740 

4  3  0.5  0 

375 

$5  5.5  0 


2.  Find  the  present  worth  and   discount,  at    8^,  of  a 
note    of   $500,    due    in    3    yr.,    and    bearing    interest    at 

6%. 


Solution. — The  amount  of 
$500  for  3  yr.,  at  6^^,  is  $590. 
The  amount  of  $1  for  3  yr., 
at  8^^,  is  $1.24.  Then,  the 
present  worth  of  $590  is  590 
^  1.24  =  $475.81;  and  the 
discount  is  $590  — $475.81  = 
$114.19. 


Prac.  17. 


OPERATION. 

500 

.0  8 

.0  6 

3 

3  0.0  0 

:24 

3 

1.0  0 

¥0 

1.2  4 

500 

590 

1.2  4  )  5  9  0  (  4  7  5. 

81 

496 

590 
4  7  5.8  1 
1  1  4.1  9 


940 

868 

720 

620 

1000 

9  92 

80 


258  HAY'S  NEW  PKACTICAL  ARITHMETIC. 

Kule. — 1.  Find  the  amount  0/  SI  for  the  given  time  at 
the  given  per  cent. 

2.  By  this  divide  the  amount  of  the  note;  this  is  the 
present  worth. 

3.  From  the  amount  of  the  note  subtract  the  present 
worth;  this  is  the  discount. 

Hem. — When  the  note  does  not  bear  interest,  of  course  the 
amount  is  the  same  as  the  face  of  the  note. 

3.  Find  the  present  worth  and  discount,  at  G^,  of  a 
note  of  $224,  due  in  2  yr.  8200,  $24. 

4.  Find  the  present  worth  and  discount,  at  6^,  of  a 
note  of  $300,  due  in  2  yr.,  and  bearing  interest  at  8^. 

$310.71,  $37.29. 

5.  Find  the  present  worth  and  discount,  at  6^,  of  a 
debt  of  $675,  due  in  5  yr.  10  mo.  $500,  $175. 

6.  Find  the  present  worth  and  discount  for  5  mo.,  at 
10%,  of  an  account  of  $368.75.  $354,  $14.75. 

7.  A  note  of  $800,  dated  September  10,  1876,  due 
January  1,  1878,  and  bearing  interest  at  6%,  was  dis- 
posed of  for  the  present  worth,  at  10%,  July  19,  1877: 
what  was  the  present  worth  at  this  date  and  the  dis- 
count? $825.65,   $37.15. 

8.  A  merchant  bought  a  bill  of  goods  amounting  to 
$775,  on  4  months'  credit:  if  money  is  worth  10%  to 
him,  what  might  he  pay  for  the  goods  in  cash?        $750. 

9.  Bought  a  bill  of  goods,  amounting  to  $260,  on  8 
•months'  credit:    if  money   is  worth    6%,  what   sum  will 

pay  the  debt  in  cash?  $250. 

10.  A  merchant  buys  a  bill  of  goods  amounting  to 
$2480:  he  can  have  4  months'  credit,  or  5%  off,  for 
cash:  if  money  is  worth  only  10%  to  him,  what  will  he 
gain  by  paying  cash?  $45.47. 


TRUE  DISCOUNT,  259 

11.  Find  the  present  worth,  at  5^,  of  a  debt  of 
S956.34,  one-third  to  be  paid  in  1  yr.,  one-third  in  2  yr., 
and  one  third  in  3  yr.  8870.60. 

12.  Omitting  the  three  days  of  grace,  what  is  the 
difference  between  the  true  discount  and  the  bank  dis- 
count of  $535,  for  1  yr.,  at  7%?  S2.45. 

13.  A  man  was  offered  $1122  for  a  house,  in  cash,  or 
$1221,  payable  in  10  mo.,  without  interest.  He  chose 
the  latter :  how  much  did  he  lose,  if  money  is  worth 
12%  to  him?  $12. 

14.  A  man  offers  to  sell  his  farm  for  $8000  in  cash,  or 
for  $10296,  payable  in  three  equal  installments  at  the 
end  of  1,  2,  and  3  years,  without  interest:  considering 
money  to  be  w^orth  10^,  what  will  be  the  gain  to  the 
buyer  by  paying  cash?  $620. 

15.  A  note  of  $2000,  dated  July  4,  1876,  due  May  1, 
1878,  and  bearing  interest  at  8^,  was  cancelled  October 
25,  1877,  by  payment  of  the  present  worth  at  6^  :  what 
was  the  present  worth,  at  this  date,  and  the  discount? 

$2223.08,    $68.92. 


Xe  HANG  EI. 


200.  1.  A  draft,  or  bill  of  exchange,  is  a  written 
order,  from  one  person  to  another,  for  a  certain  amount 
of  money. 

Rem.  1. — The  person  upon  whom  the  bill  is  drawn  is  called  the 
drawee;  the  person  in  whose  favor  it  is  drawn  is  called  the  payee. 

Rem.  2. — When  the  draft  is  to  be  paid  upon  presentation,  it  is 
called  a  sight  draft;  when  it  is  to  be  paid  at  the  end  of  a  certain 
time,  it  is  called  a  time  draft. 

2.  Exchange  is  the  method  of  making  a  payment  by 
means  of  a  draft,  or  bill  of  exchange. 

3.  There  are  two  sorts  of  exchange :  domestic  or  Inland, 
and  foreign. 

4.  Domestic  exchange  takes  place  between  localities 
in  the  same  country. 

Rem. — The  following  is  a  common  form  of  an  inland  bill  of  ex- 
change, which  is  commonly  termed  a  rfraft  or  check: 

$500.  Cincinnati,  O.,  May  1,  1877. 

At  sight,  pay  to  John  Jones,  or  order,  five  hundred 
dollars,  for  value  received,  and  charge  to  account  of 

Silas  Thompson. 
To  Charles  Smith  &  Co.,  Xew  York. 
(260) 


EXCHANGE.  261 

5.  Foreign  exchange  takes  place  between  localities  in 
different  countries. 

Rem. — The  following  is  a  common  form  of  a  foreign  bill  of 
exchange: 

£500.  Cincinnati,  O.,  May  1,  1877. 

At  sight  of  this  first  of  exchange  (second  and  third 
of  the  same  tenor  and  date  unpaid),  pay  to  Amos  Car- 
roll, or  order,  five  hundred  pounds  sterling,  for  value 
received,  and  charge  to  account  of 

Stanley  Bingham. 

To  James  Smith  &  Co.,  London. 

A  foreign  bill  of  exchange  is  usually  drawn  in  dupli- 
cate or  triplicate,  called  a  set  of  exchange;  the  different 
copies,  termed  respectively  the  firsts  second,  and  third  of 
exchange,  are  then  sent  by  different  mails,  that  miscar- 
riage or  delay  may  be  avoided.  When  one  is  paid,  the 
others  are  void. 

6.  The  acceptance  of  a  bill  of  exchange  is  the  agree- 
ment b}'  the  drawee  to  pay  it  when  due. 

KEM.-Abill  is  accepted  by  the  drawee's  writing  the  word  "ac- 
cepted," with  his  name,  across  the  face  of  the  bill;  the  bill  is  then  an 
acceptance. 

201,  To  find  the  cost  or  face  of  a  domestic  bill  of 
exchange  (Art.   170,  Eule). 

1.  What  is  the  cost  of  a  sight  draft  on  New  York  for 
$1400,  at  ^%  premium?  ^  $1407. 

2.  What  is  the  cost  of  a  sight  draft  on  Boston,  for 
S2580,  at  \%  discount?  $2567.10. 


262  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

3.  What  is  the  face  of  a  sight  draft  on  Wheeling, 
which  cost  $375.87,  at  J%   premium?  $375.40. 

4.  What  is  the  cost  of  a  sight  draft  on  Chicago,  for 
$2785,  at  1%  discount?  82778.04. 

5.  What  is  the  face  of  a  sight  draft,  which  cost 
$1852.55,  at   \\%    discount?  $1876. 

G.  What  is  the  cost  of  a  draft  on  New  Orleans  for 
$5680,  payable  in  60  days,  exchange  being  at  ^% 
premium,  and  interest  6%  ?  $5649.08. 

7.  What  is  the  cost  of  a  draft  on  IS'ew  York  for  $1575, 
payable  in  30  days,  exchange  being  at  ^^%  premium, 
and  interest  6%?  $1578.13. 

8.  The  face  of  a  draft,  payable  in  60  days,  is  $2625; 
exchange  being  at  \\%  premium,  and  interest  6%,  what 
is  the  cost  of  the  draft?  $2636.69. 


FOREIGN   EXCHANGE. 

202,     Foreign    bills    of  exchange    are    drawn    in    the 
money  of  the  country  in  which  they  are  to  be  paid. 

Rem. — The  foreign  exchange  of  the  United  States  is  chiefly  with 
Great  Britain,  France,  Germany,  and  Canada. 


ENGLISH    MONEY. 

The  unit  of  English  money  is  the  pound  sterling. 
4  farthings  make  1  penny,       marked  d. 
12  pence  "       1  shilling,  ''         s. 

20  shillings        ''       1  pound,  "        £. 

Rem. — The  usual  coins  are:  gold,  sovereign  =  1  £,  and  half  sov- 
ereign; silver,  crown  =  5  s.,  half  crown,  florin  ==2  s.,  shilling,  six- 
penny, and  three-penny;  copper,  the  penny,  half  penn}^  and 
farthing. 


EXCHANGE.  263 


FRENCH    MONEY. 

The  unit  of  French  money  is  the  franc,  marked  ft\ 
10  centimes  make  1  decime. 
10  decimes        ^'       1  franc. 

Rem. — The  usual  coins  are:  gold  pieces  for  100,  40,  20,  10,  and  5 
francs;  silce?-  pieces  for  5,  2,  1,  h,  and  \  francs;  bronze  pieces  for  10, 
5,  2,  and  1  centimes. 

GERMAN    MONEY. 

The  unit  of  German  money  is  the  mark,  which  is 
divided  into  100  pennies  (pfennige). 

Rem.- — The  usual  coins  are:  gold  pieces  for  20,  10,  and  5  marks; 
silver  pieces  for  2,  1,  and  ^  marks;  nickel  pieces  for  10,  5,  and  1 
pennies. 

Canadian  money  is  in  dollars  and  cents,  corresponding 
with  United  States  currency. 

The  par  of  exchange  is  the  comparative  value  of  the 
standard  coins  of  two  countries. 

Rem. — The  commercial  value  of  foreign  exchange  may  be  above 
or  below  the  par  value.     Quotations  are  always  in  gold. 

The  par  value  of  the  pound  is  $4.8665.  Its  quoted 
commercial  value  varies  from  $4.83  to  $4.90  gold. 

The  par  value  of  the  franc  is  $0,193.  It  is  usually 
quoted  at  about  5  fr.  14f  centimes,  equal  to  one  dollar 
gold. 

The  par  value  of  the  mark  is  $0,238.  The  commer- 
cial quotations,  always  for  four  marks,  vary  from  $0.95 
to  $0.98. 

To  find  the  cost  or  face  of  a  foreign  bill  of  exchange : 


264 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


1.  What    will    a    sight  bill    on    London,  for  £500  IOh., 
cost  in  'New  York,  exchange  being  at  $4.87? 

OPERATION. 
10    S.r=:£.5 


Solution.  — Since  20  s.  =  £1,  10  s.  r.=  £.5. 
If  £1  is  worth  $4.87,  £500.5  are  worth  $4.87 
X  500.5  =z  $2437.44. 


5  0  0.5 

4.8  7 

3~5T35 

40040 

20020 

$2437.435 


2.  How    large    a   bill    on    London    can  be  bought    for 
$1808.04,  exchange  being  at  $4.88? 


Solution.— Since  £1  is  worth  $4.88, 
as  many  pounds  can  be  bought  for 
$1808.04  as  $4.88  is  contained  times 
in  $1808.01.  It  is  contained  870 
times,  with  a  remainder.  Reduce  the 
remainder  to  shillings  by  multiplying 
by  20.  4.88  is  contained  in  the  prod- 
uct 10  times.  The  bill  will  be  for 
£870   10s. 


OPERATION. 

4.8  8  )  1  8  0  8.0  4  (  3  7  0 
1464 
3440 
341o 


244 
20 


4.88)  4  8  80(  10  s. 
488 


0 


3.  What    will  a  bill  on  London    for  £890  8s.  .cost,  ex- 
change being  at  $4.86?  $4327.34. 

4.  How    large    a   bill    on  London    can  be    bought    for 
$2130.12,  exchange  being  at  $4.88?  £436  lOs. 

5.  What  will  a  bill   on  Paris  cost   for  1290  francs,  ex- 
change being  5  fr.  15  centimes  to  $1?  $250.49. 

6.  How  large  a  bill  on  Paris  can  be.  bought  for  $1657.60, 
exchange  being  at  5  fr.  16  centimes?  8553  fr.22. 

7.  What  will  a  bill  on  Berlin  cost  for  12680  reichsmarks, 
exchange  being  $.97  per  4  reichsmarks?  $3074.90. 

8.  How  large  a  bill  on    Frankfort    can    be    bought  for 
$1470,  exchange  being  at  .98?  6000  m. 


DEFINITIONS. 

203.  1.  Insurance  Companies  agree,  for  specified  sums 
of  money,  to  pay  a  certain  amount  to  the  person  insured 
on  the  occurrence  of  a  certain  event. 

2.  The  policy  is  the  written  contract  given  by  the 
company. 

Eem. — The  persons  insured  are  called  the  policy  holders.  The 
companies  are  sometimes  stj^led  the  underwriters. 

3.  The  premium  is  the  sum  paid  to  the  company  for 
insurance. 

4.  Pire  Insurance  is  indemnity  for  a  certain  amount 
in  case  of  loss  by  fire. 

5  Marine  Insurance  is  indemnity  for  a  certain  amount 
in  case  of  loss'by  the  dangers  of  navigation. 

6.  Life  Insurance  is  an  agreement  to  pay  a  specified 
sum  at  the  death,  or  at  a  certain  time  in  the  life,  of  the 
insured. 

FIRE  AND  MARINE  INSURANCE. 

204-.  The  premium  in  fire  and  marine  insurance  is 
a  certain  percentage  of  the  amount  insured  (Art.  170, 
Eule). 

(265) 


260  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

Rkm. — Insurance  companies  will  seldom  insure  property  at  its  full 
value.     The  insurance  is  commonly  upon  §  or  ^  of  the  value. 

1.  What  is  the  cost  of  insuring  a  liouse  worth  $3375, 
at  f  of  its  value,  the  premium  being  H^  and  the  policy 
costing  $1  ? 


Solution.  —  ^  of  the  value  of  the 
house  is  $2250.  The  premium  is  U^;^ 
of  $2250,  which  is  $33.75;  adding  $1, 
the  cost  of  the  policy,  the  sum  is 
$34.75;   the  cost  of  insurance. 


OrERATION 

)  3  3  7  5 

2250 

1125 

.ou 

2 

2250 

2250 

1125 

3  3.7  5 

1.0  0 

3  4.7  5 


2.  What  is  the  cost  of  insuring  a  house  worth  $5000, 
at  f  of  its  value,  the  premium  being  ^^,  and  the  policy 
costing  S1.50?  $20.25. 

3.  A  store  is  valued  at  $12600,  and  the  goods  at 
S14400;  §  of  the  value  of  the  store  is  insured  at  f  ^  and 
^  the  value  of  the  goods  at  2%  ;  the  cost  of  the  two  pol- 
icies is  $1.25  apiece:  what  was  the  total  cost  of  in- 
surance? $209.50. 

4.  A  man  owns  a  manufactory  valued  at  $21000,  and 
a  dwelling-house  worth  $7200:  what  will  it  cost. to  in- 
sure the  manufactory,  at  ^  of  its  valuiB,  at  IJ^,  and 
the  house,  at  its  full  value,  at  f^,  the  two  policies 
costing  $1.25  each?  $23G.50. 

5.  A  man's  dwelling,  valued  at  $5600,  was  burned  ;  it 
had  been  insured,  in  a  certain  company,  20  years,  for  f 
of  its  value,  at  l^%  :  how  much  did  he  receive  from  the 
company  more  than  the  sum  total  of  the  annual  premiums? 

$2940. 

6.  A  man  secures  a  policy  of  insurance,  on  his  house, 
for    $3600,    furniture    for    $1600,    and    library  $800;    the 


INSURANCE.  267 

premium  is  ^^,  mid    cost    of  policy  $1.25:  what    is  the 
cost  of  the  insurance?  $53.75. 

7.  A  hotel  is  insured,  for  |  of  its  value,  at  1J%  ;  the 
policy  costs  $1.25  and  the  total  cost  of  insurance  is  $151.25  : 
at  what  sum  is  the  hotel  valued?  $15000. 

8.  The  cost  of  insuring  a  house  worth  $4500,  for  ^  of 
its  value,  was  $32.75 ;  the  cost  of  the  policy  was  $1.25 : 
what  was  the  per  cent  of  insurance?  ^^. 

9.  A  farmer,  with  an  insurance  of  $1000  on  his  house, 
and  $1500  on  his  barn,  in  the  Yermont  Mutual,  pays  an 
annual  assessment  of  $3.50 :  what  is  the  per  cent  of  the 
premium?  -J-^%. 

LIFE    INSURANCE. 

205.  1.  Life  Insurance  policies  are  of  two  principal 
kinds  (1)  life  policies,  (2)  endowment  policies. 

2.  A  life  policy  is  payable  at  the  death  of  the  person 
insured. 

3.  An  endowment  policy  is  payable  at  a  specified 
time,  or  at  death  if  it  occurs  within  this  time. 

Rem. — In  life  insurance  the  premium  is  commonly  a  regular 
annual  payment,  dependent,  in  amount,  upon  the  age  of  the  in- 
dividual when  he  effects  his  insurance.  The  tables  of  a  company 
show  the  annual  premium,  at  any  age,  for  $1000  of  insurance, 

1.  A  man  at  the  age  of  40  insures  his  life  for  $5000 ; 
the  company's  annual  premium  on  $1000,  for  a  life 
policy  at  this  age,  is  $31.30;  if  he  dies  at  the  age  of  70, 
how  much  money  will  he  have  paid  the  company? 

OPERATION. 

Solution. — Since  the  annual   premium  on  $1000  $31.30 

is  $31.30,  on  $5000  it  is  $31.30  X  ^  =  $1^6.50;  then,  5 

the   amount   paid,    in    30   yr.,   will   be   $156.50  X  30  15  6.50 

--=$4695.  8  0 

$r6¥5.W 


268  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  Mr.  Harris,  aged  35,  takes  out  an  endowment  policy 
in  a  life  insurance  company  for  $10000,  payable  in  10 
years ;  the  cost  of  the  annual  premium  on  $1000,  at  bis 
age,  is  $105.58 :  if  he  lives  to  receive  the  endowment,  what 
will  be  the  cost  of  the  paid-up  policy,  without  interest? 

$10553. 

3.  At  the  age  of  50,  the  cost  of  a  life  policy,  payable 
annually,  is  $47.18  on  $1000;  the  cost  of  an  endowment 
policy,  payable  in  20  years,  is  $60.45  on  $1000  ;  at  the 
end  of  20  years,  how  much  more  will  have  been  paid  on 
a  policy  of  $8000  by  the  endowment  plan  than  by  the 
life  plan?  $2123.20. 

4.  At  the  age  of  44,  a  man  insures  his  life  to  the 
amount  of  $12000  in  favor  of  his  wife;  the  company's 
annual  premium  at  this  age,  for  a  life  policy,  is  $36.46 
on  $1000 :  if  the  man  dies  after  the  payment  of  5  pre- 
miums, how  much  more  than  he  paid  out,  will  his  widow 
receive  ?  $9812.40 

5.  At  the  age  of  21,  a  young  man  takes  out  a  life 
policy  for  $5000,  upon  which  the  annual  premium  is 
$19.89  on  $1000 :  if  he  lives  to  the  age  of  75,  how  much 
will  it  cost  him  to  keep  up  his  insurance? 

$5370.30. 

6.  At  the  age  of  30,  to  secure  an  endowment  policy 
for  $1000,  payable  in  10  3  ears,  costs  an  annual  premium 
of  $104.58 ;  what  will  be  the  amount  of  the  ten  pay- 
ments  at  the  end  of  the  time,  allowing  interest  at  6%  ? 

$1390.91. 

7.  At  the  age  of  38,  a  gentleman  took  out  a  policy  for 
$6000,  on  the  life  plan,  paying  annually  $29.15  on  $1000. 
After  keeping  up  his  premiums  for  15  years,  he  suffered 
his  policy  to  lapse :  how  much  money  had  he  paid  out, 
allowing  interest  at  6%  ?  $3882.78. 


DEFINITIONS. 

206.  1.  A  tax  is  money  paid  by  the  citizens  of  a 
coimtry  for  the  support  of  government  or  for  other  pub- 
lic purposes. 

2.  A  tax  is  either  direct  or  indirect. 

3.  A  direct  tax  is  one  which  is  levied  upon  the  per- 
son or  property  of  the  citizens. 

4.  A  tax  upon  the  person  is  called  a  poll  tax;  upon 
property,  a  property  tax. 

5.  An  indirect  tax  is  one  which,  in  some  way,  is 
levied  upon  the  business  of  the  citizens. 

6.  The  taxes  of  the  United  States,  considered  in  refer- 
ence to  their  nature  and  purpose,  are  of  two  classes,  (1) 
State  and  Local  Taxes;  (2)    United  States  Revenue. 

STATE    AND    LOCAL    TAXES. 

207.  1.  The  money  for  State  and  local  purposes  arises 
chiefly  from  direct  taxation. 

Kem. — Some  revenue  accrues  to  the  State  from  the  rent  of  school 

lands,  from  licenses,  fines,  etc. 

(269) 


270  KAY'S  NEW  PKACTICAL  ARITHMETIC. 

2.  For  the  purposes  of  taxation,  property  is  ([classed 
as  Real  Estate  and  Personal  Property. 

3.  Real  Estate  is  property  which  is  fixed,  as  lands, 
houses,  etc. 

4.  Personal  Property  is  that  which  is  movable,  as 
furniture,  merchandise,  etc. 

5.  The  valuation  is  the  estimated  worth  of  the  prop- 
erty. 

# 
Rem.^ — The  vnluation  is  generally  the  basis  upon  which  to  estimate 

the  tax.  In  some  states,  however,  the  specific  tax  upon  the  polls 
must  first  be  subtracted;  in  Massachusetts,  a  sixth  part  of  the  tax  is 
assessed  upon  the  polls,  provided  it  does  not  exceed  $2  for  each  indi- 
vidual; in  Vermont,  the  basis  is  what  is  called  the  Grand  List,  which 
is  ascertained  by  dividing  the  valuation  by  100  and  adding  $2  for 
each  poll. 

6.  The  valuation  is  made  by  an  officer  called  an 
assessor. 

Rem. — This  official  makes  out  a  list  called  an  assessment  roll;  it 
contains  the  names  of  the  persons  to  be  taxed,  along  with  the  valua- 
tion of  their  property. 

208.     To  find  the  rate  of  taxation. 

The  rate  of  taxation  is  expressed  as  so  many  mills  on 
each  dollar  of  taxable  property,  or  as  such  a  per  cent 
of  it. 

1.  The  property  of  a  certain  town  is  valued  at 
$1049905  ;  there  are  483  persons  subject  to  poll-tax.  In 
a  certain  year  the  total  taxes  of  the  tow^n  are  $13323.36 ; 
the  poll-tax  being  $1.50  for  each  person,  w^hat  is  the 
rate  of  taxation  upon  the  property? 


TAXES.  271 


Solution.  —  The  poll-tax  is 
$1.50x483r=$724.50;    then,  the 

property    tax    is    $13323.36—  ^  ^^  ^  r  n 

$724.50  =  $12598.86.        Then,  '       '"^        ' 

since   the   tax   on   $1049905    is  1  '^  ^  2  3.3  6 

$12598.86,    the    tax    on    $1    is  7  2  4.5  0 

$12598.86  --  1049905  =  $0,012,  1  0  4  9  9  0  5  )  1  2  5  9  8.8  6  (.0  1  2 

12  mills,  or  1 1  ^c-  10  4  9  9  0  5 

2  099810 
2099810 


Rule. — 1.  Multiply  the  tax  on  each  poll  by  the  number 
of  polls ;  the  product  is  the  poll-tax. 

2.  From  the  total  amount  of  tax  subtract  the  poll-tax ; 
the  remainder  is  the  property  tax. 

3.  Divide  the  property  tax  by  the  valuation ;  the  quotient 
is  the  rate  of  taxation. 

Rem. — Of  course,  where  there  is  no  specific  poll-tax,  the  total 
amount  of  the  tax  is  to  be  divided  immediately  by  the  valuation. 

2.  A  tax  of  $2500  is  assessed  upon  a  certain  district 
to  build  a  school-house.  The  property  of  the  district  is 
valued  at  $618000,  and  there  are  28  persons  subject  to 
poll-tax:  if  the  poll-tax  is  $1,  what  will  be  the  rate  of 
taxation?  4  mills  on  $1,  or  1%. 

3.  Upon  a  valuation  of  $2876475  the  tax  is  $18409.44: 
there  being  no  poll-tax,  what  is  the  rate? 

6.4  mills  on  $1. 

4.  The  total  valuation  of  property  in  the  State  of 
Wisconsin,  for  j.874,  was  $421285359;  the  tax  levied  upon 
this  valuation  was  $656491.61:  what  was  the  rate  to 
the  hundredth  of  a  mill?  .  1.56  mills  on  $1. 


ITl 


KAY'S  NEW  PRACTICAL  AR1THMETI(\ 


209.     To  apportion  the  tax  among  the  tax-])ayer8. 

I.  A  tax  of  $1373.64  is  aBsessed  upon  a  village,  the 
property  of  which  is  valued  at  $748500 ;  57  persons  pay 
a  poll-tax  of  $1.25  each;  find  the  rate  of  taxation,  and 
construct  a  tax  table  to  $9000. 


TAX    TABLE. 
Rate,    1.74    mills   on    $1. 


PROP. 

TAX. 

PROP. 

TAX. 

PROP. 

TAX. 

PROP. 

TAX. 

$1 

$0,002  . 

$10 

$0,017 

$100 

$0,174 

$1000 

$  1.74 

2 

.003 

20 

.035 

200 

.348 

2000 

3.48 

3 

.005 

30 

.052 

300 

.522 

3000 

5.22 

4 

.007 

40 

.070 

400 

.696 

4000 

6.96 

5 

.009 

50 

.087 

500 

.870 

5000 

8.70 

6 

.010 

GO 

.104 

600 

1.044 

6000 

10.44 

7 

.012 

70 

.122 

700 

1.218 

7000 

12.18 

8 

.014 

80 

.139 

800 

1.392 

8000 

13.92 

9 

.016 

90 

.157 

900 

1.566 

9000 

15.66 

Rem. — In  order  to  facilitate  the  calculation  of  each  person's  tax,  it 
is  customary  to  construct  such  a  table.  It  is  not  necessary  to  carry 
it  out  in  any  column  farther  than  the  nearest  mill. 

1.  James  Turner's  property  is  valued  at  $7851,  and  he 
pays  poll-tax  for  2  persons:  what  is  his  tax? 

OPERATION. 


Solution. — By  the  table,  the  tax  on  $7000  is 
$12.18;  on  $800,  $1,392;  on  $50,  $0,087;  and  on 
$1,  $0,002;  then,  the  tax  on  $7851  is  $12.18  + 
$1,392  -f-  $0,087  +  $0,002  =r  $13.66;  this  is  his  prop- 
erty tax.  The  poll-tax  is  $1.25  X  2  =  $2.50. 
Then,  James  Turner's  tax  is  $13.66  +  $2.50  at 
$16.16. 


$     7851 


12.18  0 

1.3  9  2 

.0  8  7 

.0  0  2 

13.6  6 

2.5  0 

16.16 


UNITED  STATES  REVENUE.  273 

Explanation. — It  is  evident  that  the  operation  is  equivalent  to 
multiplying  $7851  by  the  rate,  1.74,  and  adding  the  poll-tax. 

2.  John  Brown's  property  is  valued  at  $2576,  and  he 
pays  poll-tax  for  1  person :  what  is  his  tax  ?  S5.73. 

3.  Henry  Adams'  property  is  valued  at  $9265,  and  he 
pays  poll-tax  for  3  persons:  what  is  his  tax?         $19.87. 

4.  Amos  Clarke's  property  is  valued  at  $4759,  and  he 
pays  poll-tax  for  1  person:  what  is  his  tax?  $9.53. 

5.  Emily  Wood's  property  is  valued  at  $8367 :  what 
is  her  tax?  $14.56. 

II.  The  tax  to  be  raised  in  a  city  is  $64375 ;  its  tax- 
able property  is  valued  at  $16869758  ;  find  the  rate  of 
taxation  to  thousandths  of  a  mill,  and  construct  a  tax 
table  to  $90000.  Eate  3.816  mills  on  $1. 

1.  William  Mill's  property  is  valued  at  $56875:  what 
is  his  tax?  $217.04. 

2.  Samuel  Young's  property  is  valued  at  $27543 :  what 
is  his  tax?  $105.10. 

3.  Charles  O'Neil's  property  is  valued  at  $83612 :  what 
is  his  tax?  $319.06. 

4.  Adolph  Meyer's  property  is  valued  at  $72968 :  what 
is  his  tax?  $278.45. 

5.  Louis  Ganot's  property  is  valued  at  $69547 :  what 
is  his  tax?  $265.39. 

UNITED    STATES  REVENUE. 

210.  1.  The  United  States  Revenue  arises  wholly 
from  indirect  taxation ;  it  consists  of  Internal  Revenue 
and  the  revenue  from  Duties  or  Customs. 

2.  The  Internal  Revenue  arises  from  the  sale  of  pub- 
lic lands,  from  a  tax  upon  certain  manufactures,  from 
the  sale  of  postage  stamps,  etc. 


274  KAY'S  NEW  rilACTICAL  ARITHMETIC. 

3.  Duties  or  Customs  are  taxes  on  goods  imported 
from  foreign  countries. 

INTERNAL    REVENUE. 

211.  1.  The  public  lands  are  disposed  of  at  SI. 25 
per  acre:  what  will  the  government  receive  for  a  town- 
ship containing  36  sq.  miles?  $28800. 

2.  Letter  postage  is  3  ct.  for  each  half-ounce,  or  frac- 
tion thereof:  what  is  the  postage  on  a  letter  weighing 
IJ  oz.?  9  ct. 

3.  The  postage  on  books  is  1  ct.  for  each  2  oz.,  or 
fraction  thereof:  what  is  the  postage  on  a  book  weigh- 
ing 1  lb.  5  oz.?  11  ct. 

4.  The  tax  on  ])roof  spirits  is  70  ct.  per  gallon  :  what 
is  the  tax  on  a  barrel  of  40  gallons?  $28.00. 

5.  The  tax  on  cigars  per  1000  is  $5 :  how  much  does 
this  enhance  the  price  of  a  single  cigar?  ^  ct. 

6.  The  tax  on  beer  is  $1  per  barrel  of  31  gal.  Each 
wholesale  dealer  in  malt  liquors  pays  a  special  tax  of 
$50,  and  each  retail  dealer  a  special  tax  of  $20 ;  in  a 
certain  city  there  are  12  wholesale  dealers,  250  retail 
dealers,  and  the  annual  manufacture  of  beer  is  30000 
bbl. :  what  is  the  revenue  to  government?  $35600. 

DUTIES    OR    CUSTOMS. 

212.  1.  Duties  are  of  two  kinds,  specific  and  ad 
valorem. 

2.  A  specific  duty  is  levied  upon  the  quantity  of  the 
goods. 

Rem. — In  levying  specific  duties,  allowance  is  made  ( 1 )  for  waste 
called  draft,  (2)  for  the  weight  of  the  box,  cask,  etc.^  containing  the 
goods,  called  tare.  The  waste  of  liquors,  imported  in  casks  or 
barrels,  is  called  leakage)  that  of  liquors  imported  in  bottles,  break- 
age. Gross  weight  is  the  weight  before  deducting  dry  ft  and  tare;  net 
weight  is  the  weight  after  deducting  draft  and  tare. 


DUTIES.  '  275 

3.  An  ad  valorem  duty  is  levied  upon  the  cost  of  the 
goods. 

Rem. — The  cost  of  the  goods  is  shown  by  the  foreign  invoice,  or  it 
is  determined  by  appraisement  at  the  custom-house. 

4.  Duties  must  be  paid  in  coin. 

Kem. — The  duty  is  computed  on  the  net  weight  and  on  the  total 
cost  of  the  article  in  the  foreign  country.  The  dutiable  value  upon 
which  the  duty  is  estimated,  is  always  the  nearest  exact  number  of 
dollars,  pounds,  etc. 

1.  The  gross  weight  of  a  hogshead  of  imjiorted  sugar 
is  1760  lb.;  allowing  12^%  tare,  what  is  the  duty  at 
If  ct.  per  pound?  S26.95. 

2.  A  manufacturer  imported  from  Spain  40  bales  of 
wool,  of  400  lb.  each,  tare  ^%  ;  the  cost  was  45  ct.  per 
pound:  what  was  the  duty,  at  9  ct.  per  pound  and  10^ 
ad  valorem?  $2052. 

3.  A  merchant  imported  a  case  of  glassware ;  the  cost 
of  the  ware  in  France  was  365.15  francs,  the  case  and 
charges  were  57.15  francs,  and  the  commission  5^  :  what 
was  the  duty  at  40^  in  U.  S.  money,  reckoning  the 
franc  at  19-\  ct.?  $34.40. 

Rem. — The  total  cost  being  $85.58,  the  dutiable  value  is  $86.00. 

4.  A  book-seller  imports  a  case  of  books ;  their  cost  in 
Germany  was  1317.04  marks,  case  and  charges  34.36 
marks,  and  commission  6^  :  what  was  the  duty  at  25^ 
in  U.  S.  money,  the  mark  being  estimated  at  23.8  ct.  ? 

$85.25. 

5.  A  merchant  imports  six  cases  of  woolen  cloth,  net 
weight  1500  lb. ;  the  cost  in  England  was  £500,  cases 
and  charges  £8  48.  6d.,  commission  2^^  :  what  was  the 
duty,  at  50  ct.  per  To.  and  35%  ad  valorem  in  U.  S. 
money,  estimating  the  pound  at  $4.8665?  $1637.25. 


DEFINITIONS. 

213.  1.  Ratio  is  the  relation  of  two  numbers  ex- 
pressed by  their  quotient. 

Thus,  the  ratio  of  0  to  2  is  6  —  2^3;    that  is,  6  is  3  times  2. 

Rem. — The  established  custom  in  several  departments  of  math- 
ematics makes  it  advisable  to  change  the  treatment  of  ratio  as 
given  in  former  editions  of  Ray's  Arithmetics. 

2.  The  ratio  of  two  numbers  is  indicated  by  writing 
the  sign  ( : )  between  them. 

Thus,  2  :  6  is  read  the  ratio  of  2  to  6.  • 

3.  The  two  numbers  are  styled  the  terms  of  the 
ratio. 

4.  The  first  term  is  called  the  antecedent,  and  the 
second  term  the  consequent. 

5.  6  :  2  is  3,  a  ratio  between  two  abstract  numbers. 
$6  :  S2  is  3,  a  ratio  between  two  concrete  numbers  of 
the  same  denomination.  To  find  2  yd.  :  2  ft.,  reduce 
the  2  yd.  to  ft;  6  ft.   :  2  ft.  is  3. 


A    ratio  can   not  exist   between  2  ft.  and 

(276) 


because 


KATIO.  277 

they   can   not    be    reduced    to    the    same    denomination. 
Hence, 

1st.  The  terms  of  the  ratio  may  be  either  abstract  or 
concrete. 

2d.  Whe7i  the  terms  are  concrete,  both  must  be  of  the 
same  denominatioji. 

3d.   The  ratio  is  always  an  abstract  number. 

6.  Eatios  are  either  simple  or  compound. 

7.  A  simple  ratio  is  a  single  ratio. 

Thus,  2  :  G  is  a  simple  ratio. 

8.  A  compound    ratio  consists  of  two  or  more  simple 

ratios. 

2  •  6  ^ 
Thus,  o  !  Q  MS  a  compound  ratio. 

9.  In  Eatio  three  quantities  are  considered:  (1)  the 
antecedent  J  (2)  the  co7isequent,  and  (3)  the  ratio.  Any  two 
of  these  being  given,  the  third  may  be  found. 

214.     Given  the  terms,  to  find  the  ratio. 

1.  What  is  the  ratio  of  6  to  3? 

OPERATION. 

Solution.    -The  ratio  of  0  to  3  is  6  divided  by  3,         G   :  3 
equal  to  2.  6  --  3  =  2 

2.  What  is  the  ratio  of  §  to  ^? 

OPERATION. 

Solution.  —  The  ratio  of  f  to  f^is  |  divided  by  |,  I  =  f 

or  I  multiplied  by  f ,  equal  to  |.  I  "^  f 

Rule.  —  Divide  the  antecedent  by  the  consequent. 

Rem.  —  When  the  terms  are  of  different  denominations,  they  must 
be  reduced  to  the  same  denomination. 


278 


KAYS  NEW   PKACTICAL  AKITIIMETIC. 


What  is  the  ratio  of 

3.  12  to     3?  4, 

4.  30  to     5?  6, 

5.  35  to     7?  5 

6.  56  to     8?  7 

7.  5  to  10?  ^. 

8.  7  to  21?  J. 

9.  12  to  18?  |. 

10.  15  to  20?  f 

11.  15  to  25?  f 

12.  25  to  15?  l| 


13. 

36 

to  28? 

14. 

49 

to  35? 

15. 

^ 

to     §? 

16. 

4 

to     1? 

17. 

i 

to    4? 

18. 

I 

to     i? 

19. 

n 

to     1? 

20. 

H 

to  2J? 

21. 

H 

to  2i? 

22. 

6A 

to  4f  ? 

If 
I- 


2. 
If 

^■ 


What  is  the  ratio  of 

23.  S18  to  $6? 

24.  54  days  to  9  days? 


25.  96  men  to  12  men? 

26.  221  bu.  to  17  bu.? 

27.  1  fl.  9  in.  to  3  in.? 

28.  5  yd.  1  a.  to  5  fl.  4  in.? 

215.     Given  the  ratio  and  the  consequent,  to  find  the 
antecedent. 


3. 

6. 

8. 

13. 

7. 
3. 


1.  7  is  the  ratio  of  what  number  to  4? 

OPERATION. 

Solution. — The  number  is  4  multiplied  by  7,  4  X  "  =^28 

equal  to  28. 

Rule. — Multiply  the  consequent  by  the  ratio. 


4  is  the  ratio  of  what  number  to  13? 
|-  is  the  ratio  of  what  number  to  27  ? 
-j^  is  the  ratio  of  what  number  to  52? 
2|-  is  the  ratio  of  what  number  to  24  ? 
45^  is  the  ratio  of  what  number  to  If? 


52. 
15. 

28. 
63. 
7^ 


EATIO.  279 

7.  3  is  the  ratio  of  what  to  75  ct.  ?  $2.25. 

8.  ^  is  the  ratio  of  what  to  4  lb.  8  oz.  ?      3  lb.  15  oz. 

9.  2.6  is  the  ratio  of  what  to  $4.  $10.40. 

216.  Given  the  ratio  and    the  antecedent,  to  find  the 
consequent. 

1.  5  is  the  ratio  of  45  to  what  number? 

OPERATION. 

Solution.  —  The   number   is    45   divided   by  5,         45 -=-5  =  9 
equal  to  9. 

Rule. — Divide  the  antecedent  by  the  ratio. 

2.  4  is  the  ratio  of  56  to  what  number?  14. 

3.  ^  is  the  ratio  of  42  to  what  number?  60. 

4.  2f  is  the  ratio  of  23f  to  what  number  ?  8i 

5.  7|  is  the  ratio  of  $27.20  to  what  ?  $3.60. 

217.  To  find  the  value  of  a  compound  ratio. 

6  :  2) 
1  Find  the  vahie  of  the  compound  ratio     q  •  3  f 

OPERATION. 

Solution.— The  product  of  the  antecedents  6  6X^  =  5  4 

and  9  is  54,  the  product  of  the  consequents  2  and  2X3=    6 

3  is  6;   then,  the  value  of  the  compound   ratio  is  5  4-^6=    9 
54  divided  by  6,  equal  to  9. 

Rule. — Divide  the  'product  of  the  antecedents  by  the  prod- 
uct of  the  consequents. 

Rem. — Multiplying  the  antecedents  together  and  the  consequents 

together,  evidently  reduces   the   compound  ratio   to  a   simple   one; 

6  ■  2  ^  . 
thus,  in  the  above  example  the  compound  ratio  g  "   o  >is  equivalent 

to  the  simple  ratio  54  :   6. 


280  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

Find  the  value 

2.  Of  the  compound  ratio     q  .'  ^  [  3, 

3.  Of  the  compound  ratio  o^f  !  of  [  8. 

4.  Of  the  compound   ratio  f  !  if  [  f- 

-    ^n  .1  1       *•       8  men  :     2  men.  )  ,, 

5.  Oi   the  compound  ratio  it>  j         •  24  d 

6.  Of  the  compound   ratio  ^^u^  ;  f^^^^  I  6. 

5  :  2) 

7.  Of  the  compound  ratio  7:3^  10^. 

9:5) 

218.  The  terms  of  a  ratio  correspond  to  the  terms  of 
a  fraction,  the  antecedent  to  the  numerator,  the  conse- 
quent to  the  denominator. 

Thus,  ill  2  :  3  the  ratio  is  §,  in  which  the  antecedent  2  is  the 
numerator  and  the  consequent  3  the  denominator.  Hence  (Art. 
101)  we  have  the  following 

Principles. 

I.  A  ratio  is  multiplied 

1st.   By  multiplying  the  antecedent. 
2d.  By  dividing  the  consequent. 

II.  A  ratio  is  divided 

Ist.  By  dividing  the  antecedent. 
2d.  By  multiplying  the  consequent. 

III.  A  ratio  is  not  changed 

1st.  By  midtiplying  both  terms  by  the  same  number. 
2d.  By  dividing  both  terms  by  the  same  number. 


KATIO.  281 

219.     To  reduce  a  ratio  to  its  lowest  terms. 
1.  Eeduce  16  :  24  to  its  lowest  terms. 

OPERATION. 

Solution.— The  G.  C.  D.  of  16  and  24  is  8;  divid-        8)16:24 
ing  both  terms  of  16  :  24  by  8,  it  becomes  2  :  3  (Art.  2~:     3 

218,  III,  2d). 

Rule. — Divide   both   terms  of  the   ratio  by  their  greatest 
common  divisor. 


2. 

Eeduce 

20 

3. 

Eeduce 

10 

4. 

Eeduce 

34 

5. 

Eeduce 

95 

6. 

Eeduce 

75 

7. 

Eeduce  217 

25  to  its  lowest  terms.  4  :  5. 

30  to  its  lowest  terms.  1   :  3. 

51  to  its  lowest  terms.  2  :  3. 

133  to  its  lowest  terms.  5  :  7. 

125  to  its  lowest  terms.  3:5. 

279  to  its  lowest  terms.  7  :  9. 


220.     To  clear  a  ratio  of  fractions. 


1.  Clear  1^ 


2\  of  fractions. 


Solution. — The  L.  C.  M.  of  the  denominators  2  and 
3  is  6;  multiplying  both  terms  of  K]  :  2\  by  6,  it  be- 
comes 9  :  14  (Art.  218,  III,  1st). 


operation. 
U:2^ 

6 
~^T~1"4 


Rule. — Mvltiply  both  terms  of  the  ratio  by  the  least  com- 
mon ymdtiple  of  the  denominators  of  the  fractions. 


2. 

Clear 

3| 

4f  of  fractions. 

75 

88 

3. 

Clear 

n 

lOf  of  fractions. 

45 

64 

4. 

Clear 

1 

1  of  fractions. 

15 

14. 

5. 

Clear 

61% 

9^2^.  of  fractions. 

189 

284. 

-zz:'.  z^*;— -''-O, 


DEFINITIONS. 

221.     1.  Proportion    is  an  expression  for  the  equality 
of  two  ratios. 

Thus,  2  :  4  and  3  :  6  may  form  a  proportion,  for  the  ratio  of  each 
is  \. 

2.  The  proportion  is  indicated  by  writing   :  :   between 
the  ratios. 

Thus,   2  :  4  :  :  3  :    6  is   read  2  is   to  4   as   3   is   to  6. 

3.  A  proportion  is  either  simple  or  compound. 

4.  In  a  simple  proj^ortion  both  the  ratios  are  simple. 

Thus,  2  :  4  :  :  3  :  G    is   a   simple   proportion. 

5.  In  a  compound  proportion  one   or  both    the    ratios 
are  compound. 

Thus,   o  !  4  [    •  •  r  ."  Q  [  's   a  compound   proportion. 

G.  Every  proportion  consists  of  four  terms. 

7.  The    first    and    fourth    terms    of    a    proportion    are 

called  the  extremes. 

(282) 


PROPORTION.  283 

8.  The  second  and  third  terms  of  a  proportion  are 
called  the  means. 

9.  The  last  term  is  said  to  be  a  fourth  proportional  to 
the  other  three  taken  in  order. 

Thus,  in  the  proportion  2  :  4  :  :  3  :  G,  the  extremes  are  2  and  6;  the 
means  are  4  and  3;  and  6  is  a  fourth  proportional  to  2,  4,  and  3. 

10.  When  three  numbers  form  a  proportion,  the  second 
number  is  said  to  be  a  mean  proportional  between  the 
other  two. 


Thus,  in  the  proportion  2  ;  4  :  :  4  :  8,  4   is  a  mean   proportional 
between  2  and  8. 


222.  The  operations  of  proportion  depend  upon  the 
following 

Principle. — In  every  proportion  the  product  of  the  ex- 
tremes is  equal  to  the  product  of  the  means. 

Thus,  in  the  proportion  2  :  4  :  :  3  :  6,  2  X  6  =  4  )<  3;  "i  the 
proportion  3  !  4}  :  =  5  !  ^  |  2  X  3  X  ^  X  8  ==  3  X  4  X  4  X  ^i  -^nd 
the  same  may  be  shown  for  any  other  proportion.     Hence  (36,  4), 

1st.  Tf  the  product  of  the  means  be  divided  by  one  of 
the  extremes,  the  c/uotlent  Kill  be  the  other  extreme. 

2d.  If  the  product  of  the  extremes  be  divided  by  one  of 
the  means,  the  quotient  will  be  the  other  mean. 

223.  Given  three  terms  of  a  proportion,  to  find  the 
fourth. 

1.   What  :  G   :    :  4  :  8? 


284 


RAYS  NEW  PRACTICAL  ARITHMETIC. 


Solution. — The  product  of  the  means  G  and  4,       operation. 
is  24;  then,  24  divided  by  8,  one  of  the  extremes,         GX4  =  24 


equals  8,  the  other  extreme  (222,  1st). 

:     5|, 
:   12]    • 


2.     4  :  what  :   :     '^  '     ^  I  ? 
10 


Solution. — The  product  of  tlie  ex- 
tremes, 4  X  S  X  12,  divided  by  3  X  10,  one 
of  the  means  equals  8,  the  other, mean 
(222,  2d). 


24--8. 


operation. 
2  4 


=-8 


Rule. — Divide    the   'product    of   the    terms    of   the   same 
name  by  the  other  given  term. 

Rem. — Indicate   the   operation    and   cancel  whenever   it   is   prac- 
ticable (91). 


3. 

2:8::     6  :  what? 

4. 

5  :  7  :   :   10  :  what? 

5. 

What  :  8  :   :   G   :   1()? 

6. 

5  :  what  :   :  G  :   12? 

7. 

3:7::  what  :   14? 

8. 

7   :   14  :   :  0   :  what? 

9. 

^;Jj::what:45? 

10. 

5   :     8)       .3   :4          |, 
4  :   10  )   ■   *  7   :  what   j    " 

11. 

10  :  what)    .   .  22  :   33  | 
14  :  21       j       ■  2G  :  39  1 

12. 
13. 
14. 

f  :  f  :   :  1  :  what? 
1  :  what  :   :  A  :   li? 
What  :  4§  :   :  71  :   10^? 

15. 

4  :   G   :   :  G   :  Avhat? 

24. 
14. 

3. 
10. 

G. 
18. 

20. 
21. 

15. 

A- 
I- 

3^. 
9. 


PKOrOKTlON.  285 

224.  Proportion,  when  applied  to  the  sohition  of  con- 
crete problems,  has  been  styled  "  The  Rule  of  Three,'' 
because  three  terms  are  given  to  find  the  fourth.  The 
use  of  Proportion  was  formerly  so  extensive  that  it  was 
often  called  "  The  Golden  Bide.'' 

The  solution  of  a  problem  by  proportion  consists  of 
two  parts: 

1st.  The  statement;  that  is,  the  proper  arrangement  of 
the  numbers  into  a  proportion. 

2d.    The  operation  of  finding  the  required  term. 

Rem. — In  arranging  the  numbers  in  a  proportion,  it  is  cus- 
tomary, though  not  necessary,  to  make  the  number  or  quantity  re- 
quired a  fourth  pi'oportional  to  the  other  three;  then,  the  first  three 
terms  of  the  proportion  always  are  given  to  find  the  fourth. 


I.     SIMPLE    PROPORTION. 
1.  If  2  yd.  of  cloth  cost  $4,  what  will  6  yd.  cost? 

OPERATION. 

Solution. — Since   the    number   re-  2   :   6   :    :   4   :  what? 

quired,  or  fourth  term  of  the  propor-  3 

tion,  is  dollars,  the  third  term   is  $4.  ^  X  4 

Since  the  cost  of  6  yd,  will  be  greater  "    ^ 
than   the  cost  of  2  yd.,   6  yd.  is  the 

second  term  of  the  proportion,  and  2  yd.  the  first  term.      Dividing 

the  product  of  6  and  4  by  2  (Art.  223,  Rule),   the  required  term 
is  $12. 

Rem.— In  this  example,  the  number  of  dollars  is  in  a  direct  ratio 
to  the  number  of  yards;  that  is,  the  greaier  the  number  of  yards,  the 
greater  the  number  of  dollars  they  will  cost. 


286  KAY'S  NEW  PliACTICAL  ARITHMETIC. 

2.  If  3  men  can  dig  a  cellar  in  10  days,  in  how  many 
days  can  5  men  dig  it? 

OPERATION. 

Solution. — Since  the  number  re-        5   :   3   :   :   1  0   :  what? 
quired,  or  fourth  term  of  the  propor-  2 

tion,  is  days,  the  third  term  is  10  da.  3  X  /l^  0        /, 

Since  5  men  will  dig  the  cellar  in  a  5 

less  number  of  days  than  3  men,  3 

men  is  the  second  term  of  the  proportion  and  5  n)en  the  first  term. 
Dividing  the  product  of  3  and  10  by  5  (Art.  223,  liule),  the  required 
term  is  6  da. 

Kem. — In  this  example,  the  number  of  days  is  in  an  inverse  ratio 
to  the  number  of  men;  that  is,  the  yreaicr  the  number  of  men,  the 
less  the  number  of  days  in  which  they  will  dig  the  cellar. 

Rule. — 1.  For  the  third  term,  write  that  number  which 
is  of  the  same  denomination  as  the  mimher  required. 

2.  For  the  second  term,  tcrite  the  gueat'er  of  the  two 
remaining  numbers,  when  the  fourth  term  is  to  be  greater 
than  the  third;  and  the  less,  when  the  fourth  term  is  to 
be  less  than  the  third. 

3.  Divide  the  j^^oduct  of  the  second  and  third  terms  by 
the  first;  the  quotient  will  be  the  fourth  tenn,  or  number  re- 
quired. 

3.  If  3  men  can  dig  a  cellar  in  12  days,  how  many 
men  will  dig  it  in  6  days?  (i 

4.  If  3  yd.  cloth  cost  $8,  what  cost  6  yd.?  816. 

5.  If  5  bl.  flour  cost  $30,  what  cost  3  bl.?  $18. 

6.  If  3  lb.  12  oz.  tea  cost  $3.50,  what  cost  11  lb.  4  oz.? 

$10.50. 

7.  If  2  lb.  8  oz.  of  tea  cost  $2,  w^hat  quantity  can  you 
buy  for  $5  ?  6  lb.  4  oz. 

8.  If  4  hats  cost  $14,  what  cost  10  hats?  $35. 


PROPORTION.  287 

9.  If  3  caps  cost  69  cents,  what  cost  11  caps?       $2.53. 

10.  If  4  yd.  cloth  cost  $7,  what  cost  9  yd.  ?        $15.75. 

11.  If  8  yd.  cloth  cost  $32,  what  cost  12  yd.?  $48. 

12.  If  12  yd.  cloth  cost  $48,  what  cost  8  yd.  ?  $32. 

13.  If  $32  purchase  8  yd.  of  cloth,  how  many  yards 
will  $48  buy?  12. 

14.  If  $48  purchase  12  yd.  of  cloth,  how  many  yards 
can  be  bought  for  $32?  8. 

15.' A  man  receives  $152  for  19  months'  work:  how 
much  should  he  have  for  4  months'  work?  $32. 

16.  If  8  men  perform  a  piece  of  work  in  24  days,  in 
what  time  can  12  men  perform  it?  16  days. 

17.  If  60  men  perform  a  piece  of  work  in  8  da.,  how 
many  men  will  perform  it  in  2  days?  240. 

18.  If  15  oz.  of  pepper  co^  25  ct.,  what  cost  6  lb.? 

$1.60. 

19.  If  6  gal.  of  molasses  cost  $2.70,  what  cost  26 
gal.?  $11.70. 

20.  If  5  cwt.  85  lb.  of  sugar  cost  $42.12,  what  will  35 
cwt.  25  lb.  cost?  $253.80. 

21.  If  11  yd.  of  cloth  cost  $2.50,  what  will  be  the  cost 
of  11  yd.?"'  $1,871 

22.  If  90  bu.  of  oats  supply  40  horses  6  da.,  how  long 
will  450  bu.  supply  them?  30  da. 

23.  If  6  men  build  a  wall  in  15  da.,  how  many  men 
can  build  it  in  5  da.?  18. 

24.  If  15  bu.  of  corn  pay  for  30  bu.  of  potatoes,  how 
much  corn  can  be  had  for  140  bu.  potatoes?  70  bu. 

25.  If  3  cwt.  25  lb.  of  sugar  cost  $22.60,  what  will  be 
the  cost  of  16  cwt.  25  lb.?  "  $113. 

26.  If  a  perpendicular  staff,  3  ft.  long,  cast  a  shadow 
4  ft.  6  in.,  what  is  the  height  of  a  steeple  whose  shadow 
measures  180  ft.?  120  ft. 

27.  If  a  man  perform  a  journey  in  60  da.,  traveling  9 


288  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

lir.  each  day,  in    how  many  days  can    he  perform   it  by 
traveling  12  hr.  a  day?  45. 

28.  A  merchant,  failing,  paid  60  ct.  on  each  dollar  of 
his  debts.  He  owed  A  S22()0,  and  B  $1800:  what  did 
each  receive?  A  S1320.     B  $1080. 

29.  A  merchant,  having  failed,  owes  A  $800.30;  B 
$250;  C  $375.10;  D  $500;  F  $115;  his  property,  worth 
$G12.12,  goes  to  his  creditors:  how  much  will  this  pay 
on  the  dollar?  30  ct. 

30.  If  the  4-ccnt  loaf  weigh  9  oz.  when  flour  is  $8  a 
bl.,  what  will  it  weigh  when  flour  is  $6  a  bl.?  12  oz. 

31.  I  borrowed  $250  for  G  mo. :  how  long  should  I  lend 
$300  to  compensate  the  favor?  5  mo. 

32.  A  starts  on  a  journey,  and  travels  27  mi.  a  day ;  7 
da.  after,  B  starts  and  travels  the  same  road  3G  mi.  a 
day:  in  how  many  days  will  B  overtake  A?  21. 

33.  If  William's  services  are  worth  $15§  a  mo.,  when 
he  labors  9  hr.  a  day,  what  ought  he  to  receive  for  4|^ 
mo.,  when  he  labors  12  hr.  a  day?  $91.91^. 

34.  If  5  lb.  of  butter  cost  $|,  what  cost  f  lb.  ?        $3%. 

35.  If  6  yd.  cloth  cost  $5f ,  what  cost  7f  yd.?        $6|f 
3G.  If  J  bu.  wheat  cost  $f ,  what  cost  |  bu.  ?  $^. 

37.  If  If  yd.  cloth  cost  $2^,  what  cost  2  yd.  ?  $|. 

38.  If  $29f  buy  59^  yd.  of  cloth,  how  much  will  $31J 
buy?  G2iyd. 

39.  If  .85  of  a  gallon  of  wine  cost  $1.36,  what  will  be 
the  cost  of  .25  of  a  gallon?  $0.40. 

40.  If  61.3  lb.  of  tea  cost  $44.9942,  what  will  be  the 
cost  of  1.08  lb.  ?  $0.79. 

41.  If  ^  of  a  yard  of  cloth  cost  $|,  what  will  ^V  ^^  ^ 
yard  cost?  ^^. 

42.  If  f  of  a  yard  of  velvet  cost  $4|,  what  cost  17f 
yd.?  $178.38J. 

43.  A  wheel  has  35  cogs ;  a  smaller  wheel  working  in 


PilOFOKTlON.  289 

it,  26  cogs :  in  how  many  revolutions  of  the  larger  wheel 
will  the  smaller  gain  10  revolutions?  28f. 

44.  If  a  grocer,  instead  of  a  true  gallon,  use  a  measure 
deficient  by  1  gill,  what  will  be  the  true  measure  of  100 
of  these  false  gallons?  96J  gal. 

45.  If  the  velocity  of  sound  be  1142  feet  per  sec,  and 
the  number  of  pulsations  in  a  person  70  per  min.,  what 
is  the  distance  of  a  cloud,  if  20  pulsations  are  counted 
between  the  time  of  seeing  a  flash  of  lightning  and 
hearing  the  thunder?  3  mi.  22G  rd.  2  yd.  2\  ft. 

46.  The  length  of  a  w^all,  by  a  measuring  line,  \v^s 
643  ft.  8  in.,  but  the  line  was  found  to  be  25  ft.  5. 1  in. 
long,  instead  of  25  feet,  its  sup2)osed  length :  wliat  was 
the  true  length  of  the  wall?  654  ft.  11.17  in. 


II.     COMPOUND    PROPORTION. 

225.     1.  If  2  men  earn  $20  in  5  da.,  what  sum  can  6 
men  earn  in  10  da.? 


\.i}-- 


OPERATIOIS^. 

2  0  :  what? 


Solution.— Since   the  num-  2 

ber  required  or  fourth  term  of  5 

the   proportion    is   dollars,    the  3  2 

third    term    is    $20.     Since    6  0X|0X2O 

men  can  earn   a  greater  num- 
ber of  dollars   than   2   men,   6 


?X^ 


120 


men  is  in  the  second  term  of  the  proportion  and  2  men  in  the  first 
term;  and  since  in  10  da.  a  greater  number  of  dollars  can  be  earned 
than  in  5  da.,  10  da.  is  in  the  second  term  of  the  proportion  and  5  da. 
in  the  first  term.  Dividing  the  product  of  6,  10,  and  20  by  the 
product  of  2  and  5  (Art.  223,  Rule),  the  required  term  is  $120. 

2.  If  6  men,  in  10  da.,  build  a  wall  20  ft.  long,  3  ft. 
high,  and  2  ft.  thick,  in  how  many  days  could  15  men 
build  a  wall  80  ft.  long,  2  ft.  high,  and  3  ft.  thick? 

I'rao.  19. 


290  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

OPERATION. 


10  :  what? 


Solution. —  Since    the  15 

rumber   required,  or   fourth  2  0 

term    of    the    proportion,  is  3 

days,  the  third   term    is  10  2 

days.     Since    15    men    can        94  o 

build  a  wall  in  a  less  nun,-       ^x^^X^X     ^XX P       ,  « 
ber    of  days    than    6   men  ^  ^  l^ 

6     men    is    in    the    second  ,^y/^f^y^jj/\     ^ 

term  of  the  proportion,  and  ^ 

15  men  in  the  first  term;  since  to  build  a  wall  80  Tt.  long  will  take  a 
greater  number  of  days  than  to  build  a  wall  20  ft.  long,  80  ft.  is  in 
t  second  term  of  the  proportion  and  20  ft.  in  the  first  term;  since 
to  build  a  wall  2  ft.  high  will  take  a  less  number  of  days  than  to 
build  a  wall  3  ft.  high,  2  ft.  is  in  the  second  term  of  the  proportion 
and  3  ft.  in  the  first  term;  and  since  to  build  a  wall  3  ft.  thick  will 
1  ..  e  a  greater  number  of  days  than  to  build  a  wall  2  ft.  thick,  3  ft.  is 
in  the  second  term  of  the  proportion  and  2  ft.  in  the  first  term. 
Dividing  the  product  of  6,  80,  2,  3,  and  10  by  the  product  of  15,  20, 
3,  -nd  2  (Art.  223,  Rule),  the  required  term  is  16  da. 

Rule. — 1.  For  the  third  term,  \trite  that  number  which  is 
0^  the  same  denomination  as  the  number  required. 

2.  Arrange  each  pair  of  the  numbers  forming  the  com- 
pound ratio  as  if  with  the  third  term,  they  formed  a  simple 
proportion. 

3.  Divide  the  product  of  the  numbers  in  the  second  and 
tnird  terms  by  the  product  of  the  numbers  in  the  first  term ; 
the  quotient  will  be  the  fourth  term  or  number  required. 

3.  If  a  man  travel  24  mi.  in  2  da.,  by  walking  4  hr. 
a  day:  at  the  same  rate,  bow  far  will  be  travel  in  10 
da.,  walking  8  br.  a  day?  240  mi. 

4.  If  16  men  build  18  rods  of  fence  in  12  days,  bow 
many  men  can  build  72  rd.  in  8  da.  ?  96. 

5.  If  6  men  spend  $150  in  8  mo.,  bow  mucb  will  15 
men  spend  in  20  mo.?  8^37.50. 


PARTNERSHIP.  291 

6.  I  travel  217  mi.  in  7  days  of  6  hr.  each :  how  far 
can  I  travel  in  9  days  of  11  hr.  each?  511 J  mi. 

7.  If  $100  gain  $6  in  12  mo.,  what  Hum  will  $75  gain 
in  9  mo.?  $3.37f 

8.  If  100  lb.  be  carried  20  mi.  for  20  ct.,  how  far  will 
10100  lb.  be  carried  for  $60.60?  60  mi 

9.  To  carry  12  cwt.  75  lb.  400  mi.,  costs  $57.12:  what 
will  it  cost  to  carry  10  tons  75  mi.?  $168. 

10.  If  18  men,  in  15  da.,  build  a  wall  40  rd.  long,  5 
ft.  high,  4  ft.  thick,  in  what  time  could  20  men  build  a 
wall  87  rd.  long,  8  ft.  high,  and  5  ft.  thick?         58|f  da. 

11.  If  180  men,  in  6  days,  of  10  hr.  each,  dig  a  trench 
200  yd.  long,  3  yd.  wide,  2  yd.  deep,  in  how  many  days 
can  100  men,  working  8  hr.  a  day,  dig  a  trench  180  yd. 
long,  4  yd.  wide,  and  3  yd.  deep?  24.3 


PARTNERSHIP. 

226.  1.  A  Partnership  is  an  association  of  persons 
for  the  transaction  of  business.  Such  an  association  is 
called  a  firm,  or  house,  and  each  member,  a  partner. 

2.  The  capital,  or  stock,  is  the  amount  of  money  or 
property  contributed  by  the  firm. 

3.  The  assets  are  the  amounts  due  a  firm,  together 
with  the  property  of  all  kinds  belonging  to  it. 

4.  The  liabilities  of  a  firm  are  its  debts. 

5.  The  net  capital  is  the  difference  between  the  assets 
and  liabilities. 

1.  A  and  B  engaged  in  trade;  A's  capital  was  $200; 
B's,  $300;    they  gained   $100:   find   each  partner's  share. 

Solution— The   whole  capital   is  $200  +  $300  =r  $500;  of  this  A 


292  K/iY'S  NEW  PRACTICAL  ARITHMETIC. 

owns  |ggr=|,  and  B  owns  f gg  —  |  of  the  capital;  hence,  A's  gain 
will  be  f  of  $100:=  $40,  and  B's  gain  will  be  ^  of  $100=:  $60. 

Or,  Solution.— The  whole  capital  is  $200  -f  $300  =  $500;    then, 

$500  :  $200  ::  $100  :  $40,  A's  share; 
$500  :  $300  ::  $100  :  $00,  B's  share. 

Rule. —  Take  such  part  of  the  whole  gain  or  loss,  as 
each  jmrtnefs  stock  is  part  of  the  whole  stock. ' 

Or,  Rule. — As  the  whole  stock  is  to  each  partnefs  stocky 
so  is  the  whole  gain  or  loss  to  each  partner's  gain  or  loss. 

Rem.  —This  rule  is  applicable  when  required  to  divide  a  sum 
into  parts  having  a  given  ratio  to  each  other;  as  in  Bankruptcy, 
General  Average,  6tc. 

2.  A  and  B  form  a  partnership,  with  a  capital  of 
^800 :  A's  part  is  S300 ;  B's,  $500 ;  they  gain  $232 :  what 
is  the  share  of  each  ?  A's,  $87  ;  B's,  $145. 

3.  A's  stock  was  $70;  B's,  $150;  C's,  $80;  they  gained 
$120:  what  was  each  man's  share  of  it? 

A's,  $28 ;  B's,  $60 ;  C's,  $32. 

4.  A,  B,  and  C  traded  together:  A  put  in  $200;  B, 
$400;  C,  $600:  they  gained  $427.26:  find  each  man's 
share.  A's,  $71.21;  B's,  $142.42;  C's,  $213.63. 

5.  Divide  $90  among  3  persons,  so  that  the  parts 
shall  be  to  each  other  as  1,  3,  and  5.    $10,  $30,  and  $50. 

6.  Divide  $735.93  among  4  men,  in  the  ratio  of  2,  3, 
5,  and  7.  $86.58;  $129.87;  $216.45;  $303.03. 

7.  A  person  left  an  estate  of  $22361  to  be  divided 
among  6  children,  in  the  ratio  of  their  ages,  which  are 
3,  6,  9,  11,  13,  and  17  yr. :  what  are  the  shares? 

$1137;  $2274;  $3411;  $4169;  $4927;  $6443. 

8.  Divide  $692.23  into  3  parts,  that  shall  be  to  each 
other  as  ^,  f,  and  f  $127.60;  $229.68;  $334.95. 


BANKRUPTCY.  293 


BANKRUPTCY. 

227.  A  Bankrupt  is  one  who  has  failed  to  pay  his 
debts  when  due. 

Rem. — The  assets  of  a  bankrupt  are  usually  placed  in  the  hands 
of  an  assignee,  whose  duty  it  is  to  convert  them  into  cash,  and 
divide  the  net  proceeds  among  the  creditors. 

1.  A  man,  failing,  owes  A  $175;  B,  $500;  C,  $600; 
D,  $210;  E,  $42.50;  F,  $20;  G,  $10;  his  property  is 
worth  $934.50:  what  will  be  each  creditor's  share? 

A's,  $105;  C's,  $360;  E's,  $25.50; 

B's,  $300;  D's,  $126;  F's,  $12.00;  G's,  $6. 

2.  A  man  owes  A  $234;  B,  $175;  C,  $326:  his  prop- 
erty is  worth  $492.45 :  what  can  he  pay  on  $1 ;  and 
what  will  each  creditor  get?  67  ct.  on  $1; 

A,  $156.78;  B,  $117.25;  C,  $218.42. 

3.  Mr.  Smith  failed  in  business,  owing  $37000.  His 
assignee  sold  the  stock  for  $25000,  and  charged  $4650 
for  expenses:  how  much  did  he  pay  on  the  dollar? 

55%. 

GENERAL  AVERAGE. 

228.  General  Average  is  the  method  of  apportioning 
among  the  owners  of  a  ship  and  cargo,  losses  occasioned 
by  casualties  at  sea. 

1.  A,  B,  and  C  freighted  a  ship  with  108  tuns  of 
wine.  A  owned  48,  B  36,  and  C  24  tuns ;  they  Were 
obliged  to  cast  45  tuns  overboard  :  how  much  of  the  loss 
must  each  sustain?  A,  20;  B,  15;  C,  10  tuns. 


294  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  From  a  ship  valued  at  SI 0000,  with  a  cargo  valued 
at  $15000,  there  was  thrown  overboard  goods  valued  at 
$1125:  what  %  was  the  general  average,  and  what  was 
the  loss  of  A,  whose  goods  were  valued  at  $2150? 

General  average,  4J%;  A's  loss,  $9G.75. 


PARTNERSHIP  WITH    TIME. 

229.  1.  A  and  B  built  a  wall  for  $82;  A  had  4  men 
at  work  5  days,  and  B  3  men  7  days :  how  should  they 
divide  the  money? 

Solution. — The  work  of  4  men  5  da.  equals  the  work  of  4  X  S» 
or  20  men  1  da.;  and  the  work  of  3  men  7  da.,  equals  the  work  of 
3X7,  or  21  men  1  da.;  it  is  then  required  to  divide  $82  into  two 
parts,  having  the  same  ratio  to  each  other  as  20  to  21;  hence,  A's 
part  is  If  of  $82  =  $40;   B's  part  is  f|  of  $82  :=r  $42. 

2.  A  put  in  trade  $50  for  4  mo.;  B,  $60  for  5  mo.; 
they  gained  $24:  what  w^as  each  man's  share? 

Solution.— $50  for  4  mo.  equals  $50X4  =  $200  for  1  mo.;  and 
$60  for  5  mo.  equals  $60X5  =  $300  for  1  mo.  Hence,  divide  $24 
into  two  parts  having  the  same  ratio  as  200  to  300,  or  2  to  3.  This 
gives  A  I  of  $24  =  $9.60,  and  B  |  of  $24  =  $14.40. 

Rule. — Multiply  each  partnefs  stock  by  the  time  it  was 
employed;  then  take  such  part  of  the  gain  or  loss  as  each 
partner's  product  is  part  of  the  sum  of  all  the  products. 

3.  A  and  B  hire  a  pasture  for  $54 :  A  pastures  23 
horses  27  da.;  B,  21  horses  39  da.:  Avhat  will  each  pay? 

A,  $23.28|;  B,  $30.71i. 

4.  A  put  in  $300  for  5  mo. ;  B,  $400  for  8  mo. ;  C, 
$500  for  3  mo.:  they  lost  $100;  find  each  one's  loss. 

A's,  $24.19i|;  B's,  $51.61^;  C's,  $24.19^. 


EQUATION  OF  PAYMENTS.  295 

5.  A,  B,  and    C  hire  a  pasture  for  $18.12:  A  pastures 

6  cows    30    da. ;    B,  5    cows    40    da. ;    C,  8    cows  28  da. : 
what  shall  each  pa}^?  A,  S5.40;  B,  $6;  C,  S6.72. 

6.  Two  men  formed  a  partnership  for  16  mo. :  A  put 
in,  at  first,  $300,  and,  at  the  end  of  8  mo.,  $100  more ; 
B  put  in,  at  first,  $600,  but,  at  the  end  of  10  mo.,  drew 
out  $300;    they  gained  $442.20:    find  each  man's  share. 

A's,  $184.80;  B's,  $257.40. 

7.  A  and  B  are  partners :  A  put  in  $800  for  12  mo., 
and  B,  $500.     What  sum  must  B  put  in  at  the  end  of 

7  mo.  to  entitle   him  to  half  the  year's  profits?         $720. 

EQUATION  OF  PAYMENTS. 

230.  Equation  of  payments  is  the  method  of  finding 
the  mean  or  average  time  of  making  two  or  more  pay- 
ments, due  at  different  times. 

1.  A  owes  B  $2,  due  in  3  mo.,  and  $4,  due  in  6  mo. : 
at  wlvdt  period  can  both  '  sums  be  paid  so  that  neither 
party  will  be  the  loser? 

Solution.  —  The  interest  on  $2  for  3  mo.  equals  operation. 

the  interest  on  $1  for  3  X  2  =  6  mo.;  the  interest  2  X  ^  ==     ^ 

on  $4  for  6  mo.  equals  the  interest  on  $1  for  6  X  4  4  X  6  ==  2  4 

==  24  mo.;    then,   the  interest  on   $2  +  $4  =  $6  6              y^O 

equals  the  interest  on  $1  for  6  mo.  -j-  24  mo.  =  30  5 
mo.;  hence,  $6  must  be  on  interest  30  ^-  6  =  5  mo. 

Rule. — 1.  Multiply  each  payment  by  the  time  to  elapse 
till  it  becomes  due. 

2.  Divide  the  sum  of  the  products  by  the  sum  of  the  pay- 
ments;  the  quotient  will  be  the  equated  time. 

Rem. — When  one  of  the  payments  is  due  on  the  day  from  which 
the  equated  time  is  reckoned,  its  product  is  0;  but,  in  finding  the 
sum  of  the  payment,  this  must  be  added  with  the  others. 


2m  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  A  owes  B  $2,  due  in  4  mo.,  and  $6,  due  in  8  mo.; 
find  the  average  time  of  paying  both  sums.  7  mo. 

3.  A  owes  B  ^8,  due  in  5  mo.,  and  $4,  due  in  8  mo. : 
find,  the  mean  time  of  payment.  6  mo. 

4.  A  buys  $1500  worth  of  goods ;  S250  are  to  be  paid 
in  2  mo.,  $500  in  5  mo.,  $750  in  8  mo.:  find  the  mean 
time  of  payment.  6  mo. 

5.  A  owes  B  $300;  1  third  due  in  6  mo.;  1  fourth  in 
8  mo.;  the  remainder  in  12  mo.:  what  is  the  average 
time  of  payment?  9  mo. 

6.  I  buy  $200  worth  of  goods ;  1  fifth  to  be  paid  now  ; 
2  fifths  in  5  mo. ;  the  rest  in  10  mo. :  what  is  the  average 
time  of  paying  all?  6  mo. 

231.  In  finding  the  Average  or  Mean  time  for  the  pay- 
ment of  several  sums  due  at  difixirent  times,  any  date 
may  be  tal^en  from  which  to  reckon  the  time. 

1.  A  merchant  buys  goods  as  follows,  on  60  days 
credit:  May  1st,  1848,  $100;  June  15th,  $200:  what  is  the 
average  time  of  payment?  Ju\y  30th. 

Solution. — Counting  from  May  1,  it  is  operation. 

60  days  to  the  time  of  the  first  payment,  $100  X    ^>0—    6000 

and  105  days  to  that  of  the  second;  then,  $200  X  105=^21000 

the  equated  time  is  90  days  from  May  1st,  $300                 )  27000 

that  is,  July  30th.  ^90 

2.  I  bought  goods  on  90  days  credit,  as  follows:  April 
'2d,  1853,  $200;  June  Ist,  $300:  what  is  the  average  time 
of  payment?  Aug.  6th. 

3.  A  merchant  bought  goods  as  follows:  April  6,  1876, 
on  3  mo.,  $1250;  May  17,  1876,  on  4  mo.,  $4280;  June 
21,  1876,  on  6  mo.,  $675:  what  is  the  average  time  of 
payment?  Sept.  12,  1876. 


AVERAGE.  297 


AVERAGE. 

232.  Average  is  the  method  of  finding  the  mean  or 
average  price  of  a  mixture,  when  the  ingredients  com- 
posing it,  and  their  jDrices,  are   known. 

1.  I  mix  4  pounds  of  tea,  worth  40  ct.  a  lb.,  with  6 
lb.,  worth  50  ct.  a  lb. :  what  is  1  lb.  of  the  mixture 
worth  ? 

OPERATION. 

Solution. — 4  lb.  at  40  ct.  per  lb.  are  worth  4  X  -40  =  1.60 

$1.60,  and  6  lb.  at  50  ct.   are    worth    $3.00;  6  X  -50  rrr  3.00 

then,  4  -f-  6  =:^  10  lb.  are  worth  $4.60;  hence,  10  )  4.60 

"746 

Rule. — Divide  the  whole  cost  by  the  whole  number  of 
ingredients ;   the  quotient  will  be  the  average  or  mean  price. 

2.  Mix  6  lb.  of  sugar,  at  3  ct.  a  lb.,  with  4  lb.,  at  8 
ct.  a    lb.,  what   will    1    lb.  of  the    mixture   be    worth? 

5  ct. 

3.  Mix  25  lb.  sugar,  at  12  ct.  a  lb.,  25  lb.,  at  18  ct., 
and  40  lb.,  at  25  ct. :  what  is  1  lb.  of  the  mixture  worth? 

19|  ct 

4.  A  mixes  3  gal.  water,  w^ith  12  gal.  wine,  at  50  ct.  a 
gal.:  what  is  1  gal.  of  the  mixture  worth?  40  ct. 

5.  I  have  30  sheep :  10  are  worth  $3  each ;  12,  $4  each  ; 
the  rest,  $9  each :  find  the  average  value.  S5. 

6.  On  a  certain  day  the  mercury  in  the  thermometer 
stood  as  follows:  from  6  till  10  A.  M.,  at  03°^  from  10 
A.  M.  till  1  P.  M.,  70°  ;  from  1  till  3  P.  M.,  75°  ;  from 
3  till  7  P.  M.,  73°  ;  from  7  P.  M.  till  6  A.  M.  of  the  next 
da}^,  55°  :  what  was  the  mean  temperature  of  the  da}^, 
from  sunrise  to  sunrise?  62X°. 


DEFINITIONS. 

233.  1.  Involution  is  the  multiplication  of  a  number 
into  itself  one  or  more  times. 

2.  A  power  is  the  product  obtained  by  involution. 

3.  The  first  power  is  the  number  itself 

4.  The  second  power,  or  square,  is  the  product  ob- 
tained by  taking  the  number  twice  as  a  factor. 

Thus,  2  X  2  ==  4,  is  the  second  power  or  square  of  2. 

5.  The  third  power,  or  cube,  is  the  product  obtained 
by  taking  the  number  three  times  as  a  factor. 

Thus,  2  X  2  X  2  =  8  is  the  third  power  or  cube  of  2. 

Rem. — The  second  power  is  called  the  square,  because  the  area  of  a 
square  is  the  product  of  two  equal  factors  (Art.  68).  The  third 
power  is  called  the  cube,  because  the  solid  contents  of  a  cube  is  the 
product  of  three  equal  factors  (Art.  70). 

6.  The  higher  powers  of  a  number  are  denominated 
respectively  the  fourth  power^  fifth  power ^  sixth  power ^  etc. 

Thus,  2  X  2  X  2  X  2  —  16,   is  the  fourth  power  of  2; .  2  X  2  X  2 
X  2  X  2  ==  32,  is  the  fifth  power  of  2;  2X2X2X2X2X2  =6  4, 
is  the  sixth  power  of  2,  etc. 
(298) 


INVOLUTION. 


299 


7.  The  exponent  is  a  number  denoting  the  power  to 
which  the  given  number  is  to  be  raised. 

Thus,  in  3 2,  read  3  square,  the  2  denotes  the  square  of  3;  hence, 
32=9.  In  53,  reads  cube,  the  3  denotes  the  cube  of  5;  hence, 
6^  z=  125.     7*  is  read  1  fourth  power  j  9^,  %  Jifth  power,  etc. 

234.     To  raise  a  number  to  any  power. 


1.  Find  the  cube  of  75. 

Solution. — 75  multiplied  by  75  is 
5625;  this  is  the  square  of  75.  5625 
multiplied  by  75  is  421875;  this  is  the 
cube  of  75. 


OPERATION. 


75 

75 

375 

525 

5625 


5625 

7_5 

28125 
39375 
421875 


Rule. —  Obtain  a  product  in  which  the  number  is  taken  as 
a  factor  as  many  times  as  there  are  units  in  the  exponent 
of  the  power. 


2. 

Find  the  square  of  65. 

4225. 

3. 

Find  the  cube  of  25. 

15625. 

4. 

Find  the  fourth  power  of  12. 

20736. 

5. 

Find  the  fifth  power  of  10. 

100000. 

6. 

Find  the  sixth  power  of  9. 

531441. 

7. 

Find  the  eighth  power  of  2. 

256. 

8. 

Find  the  square  of  f . 

|. 

9. 

Find  the  cube  of  |. 

If- 

10. 

Find  the  fourth  power  of  f . 

m- 

11. 

Find  the  fifth  powder  of  |. 

2%- 

12. 

Find  the  square  of  16i. 

2721 

13. 

Find  the  cube  of  12f 

19531-. 

14. 

Find  the  fourth  power  of  .25 

.00390625 

15. 

143  =what? 

2744. 

16. 

194=  what? 

130321. 

17. 

(21)5  zz^  what? 

69A%- 

^ 


EVOLUTION 


DEFINITIONS. 

235.     1.  Evolution  is  the  process  of  resolving  a  num- 
ber into  two  or  more  equal  factors. 

2.  A   root    of  a   number  is   one    of  the  two  or  more 
equal  factors. 

3.  The  square  root  of  a  number  is  one  of  two  equal 
factors. 

Thus,  3  is  the  square  root  of  9;  for  9  =  3  X  3. 

4.  The  cube  root   of  a  number  is  one  of  three  equal 

factors. 

Tlius,  3  is  the  cube  root  of  27;  for  27  =r  3  X  3  X  3. 

5.  The   higher   roots    of  a    number    are    denominated 
respectively  the  fourth  root,  fifth  root,  etc. 

Thus,  3  is  the  fourth  root  of  81;  for  81  =  3  X  3  X  ^  X  3.     3  is  the 
fifth  root  of  243;  for  243  =  3  X  3  X  3  X  3  X  3. 

6.  The    radical    sign    |/         placed    before    a    number 
shows  that  its  root  is  to  be  extracted. 

7.  The    index    is    a    number   placed  above  the  radical 
sign  to  show  the  number  of  the  root. 

(300) 


EVOLUTION. 


301 


Rem. — It  is  customary,  however,  to  omit  2,  the  index  of  the 
square  root. 

Thus,  VH^  is  read  the  square  root  of  25;  hence,  l/^=r5. 
^"^  is  read  the  cube  root  of  27;  hence,  f~2T—Z.  ^T6  is 
read   the   fourth   root   of   16;    hence,   Vl6:=2. 

8.  A  perfect  power  is  one  whose  root  can  be  ob- 
tained exactly. 

Thus,  25  and  if  are  perfect  squares;  27  and  ^V  are  perfect  cubes; 
16  and  Jg  are  perfect  fourth  powers. 

9.  The  squares  and  cubes  of  the  first  ten  numbers  are 
exhibited  in  the  following 


TABLE. 


Numbers. 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

Squares. 

1 

4 

9 

16 

25 

36 

49 

64 

81 

100 

Cubes. 

1 

8 

27 

64 

125 

216 

343 

512 

729 

1000 

Rem. — The  numbers  in  the  first  line  are  the  square  roots  of  the 
corresponding  numbers  in  the  second  line,  and  the  cube  roots  of  those 
in  the  third  line. 


10.     An    imperfect    power   is   one  whose   root  can  be 
obtained  only  approximately. 

Thus,    ]/Tz=:  1.41421 +. 


302  RAY'S  NEW  PRACTICAL  ARITHMETIC. 


SQUARE   ROOT. 

236.  To  find  the  number  of  figures  in  the  square 
root. 

1.  The  square  root  of  1  is  1,  and  the  square  root  of 
100  is  10  (Art.  235,  9,  Table);  between  1  and  100  are  all 
numbers  consisting  of  one  or  two  figures,  and  between 
1  and  10  are  all  numbers  consisting  of  one  figure ;  there- 
fore, 

When  a  number  consists  of  one  or  two  figures^  its  square 
root  consists  of  one  figure. 

2.  The  square  root  of  100  is  10,  and  the  square  root 
of  10000  is  100;  between  100  and  10000  are  all  num- 
bers consisting  of  three  or  four  figures,  and  between  10 
and  100  are  all  numbers  consisting  of  two  figures ;  there- 
fore, 

When  a  number  consists  of  three  or  four  figures^  its 
square  root  consists  of  tico  figures. 

3.  In  like  manner  it  may  be  shown  that, 

When  a  number  consists  of  five  or  six  figures,  its  square 
root  consists  of  three  figures. 

And  so  on  ;  therefore, 

1st.  If  a  number  be  pointed  off  into  periods  of  tivo  figures 
each,  the  number  of  periods  will  be  the  same  as  the  number 
of  figures  in  the  square  root. 

2d.  The  square  of  the  units  will  be  found  in  the  first 
period,  the  square  of  the  tens  in  the  second  period,  the 
square  of  the  hundreds  in  the  third  period,  etc. 


SQUAKE  ROOT.  303 

237.  To  point  off  a  number  into  j^eriods  of  two 
figures  each. 

1.  Point  off  368425.  368425. 

2.  Point  off  6.843256.  6.843256 

3.  Point  off  83751.42963.  83751.429630 

Rule. — Place  a  point  over  the  order  units,  and  then  over 
every  second  order  from  units  to  the  left  and  to  the  right. 

Rem.  1. — The  first  period  on  the  left  of  the  integral  part  of  the 
number  will  often  contain  but  a  single  figure. 

Rem.  2. — When  the  first  period  on  the  right  of  the  decimal  part 
contains  but  a  single  figure,  a  cipher  must  be  annexed  to  complete 
the  period. 

4.  Point  off  864326 ;  4.758462 ;  7584.3769. 

5.  Point  off  97285.46138 ;  75300 ;  .046827 ;  .0625 ;  .625. 

238.     To  extract  the  square  root  of  a  number. 
1.  Extract  the  square  root  of  256. 

OPERATION. 

Solution. — Point  off  256  into  periods  of  two  2  5  6(16 

figures  each  by  placing  a  point  over  6  and  2  1 

(Art.  237,  Rule).  26)156 

156 

The  largest  square  in  2  (Art.  235,  9,  Table)  is  1;  its  root  is  1; 
place  the  root  1  on  the  right  and  subtract  the  square  1  from  2; 
the  remainder  is  1,  to  which  bring  down  the  next  period  56. 

Double  the  root  1  and  place  the  result  2  on  the  left  of  156  for  a 
trial  divisor.  Find  how  many  times  2  is  contained  in  15  (making 
allowance  for  subsequent  increase  of  the  trial  divisor);  the  result  is 
6;  place  6  in  the  root  on  the  right  of  1  and  also  on  the  right  of  2, 
the  trial  divisor;  then  26  is  the  complete  divisor.  Multiply  26  by  6 
and  subtract  the  product  156  from  156;  the  remainder  is  0.  There- 
fore, 256  is  a  perfect  square,  and  its  square  root  is  16. 


304 


HAYS  NEW  PRACTICAL  ARITHMETIC. 


GEOMETRICAL  EXPLANATION. 


■  1  1  1  II 

D 

-  6X6=36 

1    1    1    1    1    1    1    1    I— 

B 

10X6=60 

1        C 

-  10X6=60 

A. 
10X10=100 

After  finding  that  the  sq.  root 
of  the  given  number  will  con- 
tain two  places  of  figures  (tens 
and  units),  and  that  the  figure 
in  tens'  place  is  1  (ten),  form  a 
square  figure  (A)  10  in.  on  each 
side,  which  contains  (Art.  67) 
100  sq.  in.;  taking  this  sum 
from  the  whole  number  of 
squares,  156  sq.  in.  remain, 
which  correspond  to  the  num- 
ber, 156,  left  after  subtracting 
above. 

It  is  obvious  that  to  increase  the  figure  A,  and  at  the  same  time 
preserve  it  a  square,  both  length  and  breadth  must  be  increased 
equally;  and,  since  each  side  is  10  in.  long,  it  will  take  twice  10,  that 
is,  20  in.,  to  encompass  two  sides  of  the  square  A.  For  this  reason, 
10  is  doubled  in  the  numerical  operation. 

Now  determine  the  breadth  of  the  addition  to  be  made  to  each 
side  of  the  square  A.  After  increasing  each  side  equally,  it  will 
require  a  small  square  (D)  of  the  sa.vie  breadth  as  each  of  the  fig- 
ures B  and  C,  to  complete  the  entire  square;  hence,  the  superficial 
contents  of  B,  C,  and  D,  must  be  equal  to  the  remainder,  156. 
Now  their  contents  are  obtained  by  multiplying  their  length  by 
their  breadth. 

Then  the  figure  in  the  units'  place — that  is,  the  breadth  of  B  and 
C — must  be  found  by  trial,  and  it  will  be  somewhat  less  than  the 
number  of  times  the  length  of  B  and  C  (20)  is  contained  in  the  re- 
mainder (156).  20  is  contained  in  156  more  than  7  times;  let  us 
try  7:  7  added  to  20  makes  27  for  the  whole  length  of  B,  C,  and 
D,  and  this,  multiplied  by  7,  gives  189  for  their  superficial  contents; 
this  being  more  than  156,  the  breadth  (7). was  taken  too  great. 
Next,  try  6  for  the  length  and  breadth  of  D;  adding  6  to  20  gives 
26  for  the  length  of  B,  C,  and  D;  multiplying  26  by  the  breadth 
(6)  gives  156  for  the  superficial  contents  of  B,  C,  and  D. 

Hence,  the  square  root  of  256  is  16;  or,  when  256  sq,  in.  are 
arranged  in  the  form  of  a  square,  each  side  is  16  inches. 


SQUAKE  KOOT.  305 

2.  Extract  the  8qiiare  root  of  758.436. 

OPERATION. 

Solution. — Point     off     758.43G  ^ro^o^.  A/o-rro  , 

.     T  ,»  ^  '  ^  "•'*  <^0U(   Z  I  .O  O  -4- 

into    periods   of   two   ngures   each  . 

by   placiiiijj    a    point    over   8    and  

then    over    7    to    the    left,    and    3  4  /  j  .i  o » 
and    0    to    the    right    (Art.  237,  ^ '^  ^ 


Rule),     Then   find   the    figures  of  5  45)2943 

the  root  as  in  Ex.  1.     The  last  re-  2  7  2  5 


mainder  is  5351.    Therefore,  758.43G  5503)21860 

is    an    imperfect    square,    and    its  16  5  0  9 

square  root  is  27.53  -f.  5  3  5  1 

Rem. — By  bringing  down  one  or  more  periods  of  decimal  ciphers, 
the  operation  might  be  continued  to  any  required  number  of  decimal 
places  in  the  root. 

3.  Extract  the  square  root  of  ||f . 

Solution. — The  square  root  of  the  numerator  256  is  16,  and 
the  square  root  of  the  denominator  625  is  25  (Ex.  1);  then,  the 
square  root  of  ||f  is  if. 

4.  Extract  the  square  root  of  f . 

Solution. — |  reduced  to  a  decimal  is  .375,  The  square  root  of 
.375,  to  five  decimal  places,  is  .61237  (Ex.  2);  then,  the  square  root 
of  f  is  .61237  -f . 

Rule. — 1.  Point  off  the  given  number  into  periods  of  two 
figures  each. 

2,  Find  the  greatest  square  in  the  first  period  on  the  left ; 
place  its  root  on  the  rights  like  a  quotient  in  division;  sub- 
tract the  square  from,  the  period,  and  to  the  remainder  bring 
down  the  next  period  for  a  dividend. 

3.  Double  the  root  found,  and  place  it  on  the  left  of  the 
dividend  for  a  trial  divisor.  Find  how  many  times  the 
trial  divisor  is  contained  in  the  dividend,  exclusive  of  the 
right  hand,  figure;  place  the  quotient  in  the  root^  and  also 
on  the  right  of  the  trial  divisor. 


306 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


4.  Multiply  the  complete  divisor  by  the  last  figure  of  the 
root;  subtract  the  product  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

5.  Double  the  ivhole  root  found,  for  a  new  trial  divisor, 
and  continue  the  operation  in  the  same  manner  until  all  the 
periods  are  brought  down. 

Rem.  1. — When  the  luiniher  is  an  imperfect  square,  the  operation 
may  be  continued  to  any  required  number  of  decimal  places  in  the 
root  by  bringing  down  periods  of  decimal  ciphers  (Ex.  2). 

Rem.  2. — To  extract  the  square  root  of  a  common  fraction:  (1) 
when  both  terms  are  perfect  squares,  extract  the  square  root  of  the 
numerator  and  then  of  the  denominator  (Ex.  3);  (2)  wh(;n  both 
terms  are  not  perfect  squares,  reduce  the  fraction  to  u  decimal  and 
extract  the  square  root  of  the  decimal  ( Ex.  4 ). 

Extract  the  square  root  of 


5. 

529. 

23. 

17. 

915.0625. 

30.25. 

6. 

625. 

25. 

18. 

.0196. 

.14 

7, 

6561. 

81. 

19. 

1.008016. 

1.004 

g. 

56644. 

238. 

20. 

.00822649. 

.0907 

9. 

390625. 

625. 

21. 

TT^- 

2^. 

10. 

1679616. 

1296. 

22. 

t¥A- 

H. 

11. 

5764801. 

2401. 

23. 

30i 

^. 

12. 

43046721. 

6561. 

24. 

10. 

3.16227 -f 

13. 

987656329. 

31427. 

25. 

2. 

1.41421-f 

14. 

289442169. 

17013. 

26. 

3- 

.81649  + 

15. 

234.09. 

15.3. 

27. 

6f. 

2.52982  -f 

16. 

145.2025. 

12.05. 

28. 

384f 

19.61049  + 

239.     To  extract  the   square   root  of  a  perfect  square 
by  factoring. 


1.  Exti-Mct  llie  square  root  of  441. 


SQUAKE  KOOT.  307 

Solution.  -441  =3X  3X  7X  7;     hence,    v^  44l  =  3  X  7  =  21. 

Rule. — Besolve   the   number  into    its   prime  factors,  and 
find  the  product  of  one  of  each  two  equal  factors. 

Extract  the  square  root  of 


2. 

16. 

4. 

6. 

400. 

20. 

3. 

36. 

6. 

7. 

1764. 

42. 

4. 

100. 

10. 

8. 

5184. 

72. 

5. 

225. 

15. 

9. 

3025. 

55. 

240,     Given    two    of  the    sides   of  a  right-angled  tri- 
ani^C^e  to  find  the  third  side. 


] .  A  triangle  is  a  plane  figure 
bounded  by  three  straight  lines,  called 
its  sides. 


Thus,  D  E  F  is  a  triangle;    its  sides  are  D  E,  E  F,  and  D  F. 


2.  When  one  of  the  sides  is  perpendicular  to  another, 
they  form  a  right-angle,  and  the  triangle  is  called  a 
right-angled  triangle. 

A 
Thus,  in  the  triangle  A  B  G,  the  side  A  C 
being  perpendicular  to  the  side  B  C,  they  form 
a  right-angle  at  C;   hence,  ABC   is  a  right- 
angled  triangle. 

3.  The  side  opposite  the  right-angle  is  called  the 
hypotenuse ;  the  other  two  sides,  the  base  and  the  per- 
pendicular. 

Thus,  in  A  B  C,  A  B  is  the  hypotenuse,  B  C  the  base,  and 
A  C   the   perpendicular. 


308 


KAY'S  NEW  PRACTICAL  ARITHMETIC. 


4.  Proposition. —  The  square  described  on  the  hypotenuse 
of  a  right-angled  triangle  is  equal  to  the  sum  of  the  square.^ 
described  on  the  other  two  sides. 


Draw  a  right-angled  triangle,  ABC, 
with  the  side  B  C  4  in.,  and  the  side 
A  C  3  in.;  then,  the  side  A  B  will  be 
5  in.  Describe  a  square  on  each  side  of 
the  triangle,  and  divide  each  square  into 
smaller  squares  of  1  in.  to  the  side.  Then, 
the  square  described  on  A  B  will  contain 
25  square  inches,  and  the  two  squares 
described  on  B  C  and  A  C  will  contain 
la-j-  9  =;=  25  square  inches. 


I    I     c 


5.  From  this  proposition  we  deduce  the  following 

Rules. — Ist.  To  find  the  hypotenuse;  To  the  square  of 
the  base  add  the  square  of  the  perpendicular,  and  extract 
the  square  root  of  the  sum. 

2d.  To  find  the  base  or  the  perpendicular;  From  the 
square  of  the  hypotenuse  subtract  the  square  -of  the  other 
given  side,  and  extract  the  square  root  of  the  difference. 

1.  The  base  and  perpendicular  of  a  right-angled  tri- 
angle are  30  and  40:  what  is  the  hj'potenuse?  50. 

2.  The  hypotenuse  of  a  right-angled  triangle  is  100, 
and  the  base  60:  what  is  the  perpendicular?  80. 

3.  A  castle  45  yd.  high  is  surrounded  by  a  ditch  60 
yd.  wide :  what  length  of  rope  will  reach  from  the  out- 
side of  the  ditch  to  the  top  of  the  castle?  75  yd. 

4.  A  ladder  60  ft.  long  reaches  a  window  37  ft.  from 
the  ground  on  one  side  of  the  street,  and,  without  mov- 
ing it  at  the  foot,  will  reach  one  23  ft.  high  on  the  other 
side:  find  the  width  of  the  street.  102.64+  ft. 


CUBE  HOOT.  309 

5.  A  tree  140  ft.  high  is  in  the  center  of  a  circular 
island  100  ft.  in  diameter;  a  line  600  ft.  long  reaches 
from  the  top  of  the  tree  to  the  further  shore :  what  is 
the  breadth  of  the  stream,  the  land  on  each  side  being 
of  the  same  level  ?  533.43  -f-  ft. 

6.  A  room  is  20  ft.  long,  16  ft.  wide,  and  12  ft.  high : 
what  is  the  distance  from  one  of  the  lower  corners  to 
the  opposite  upper  corner?  28.28 -f  ft. 

241.  Given  the  area  of  a  square  to  find  its  side 
(Art.  67). 

Biile. — Extract  the  square  root  of  the  area. 

1.  The  area  of  a  square  field  is  6241  sq.  rd. :  what  is 
the  length  of  one  side?  79  rd. 

2.  The  surface  of  a  square  table  contains  8  sq.  ft.  4 
sq.  in.:  what  is  the  length  of  one  side?  2  ft.  10  in. 

3.  The  area  of  a  circle  is  4096  sq.  yd. :  what  is  the 
side  of  a  square  of  equal  area?  64  yd. 

4.  A  square  field  measures  4  rd.  on  each  side :  what 
is  the  length  of  the  side  of  a  square  neld  which  contains 
9  times  as  many  square  rods?  12  rd. 

5.  What  is  the  length  of  one  side  of  a  square  lot  con- 
taining 1  acre?  208.71+  ft. 

CUBE  ROOT. 

242.  To  find  the  number  of  figures  in  the  cube  root. 

1.  The  cube  root  of  1  is  1,  and  the  cube  root  of  1000 
is  10  (Art.  235,  9,  Table);  between  1  and  1000  are  all 
numbers  consisting  of  one,  two,  or  three  figures,  and 
between  1  and  10  are  all  numbers  consisting  of  one 
figure ;  therefore, 


310  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

When  a  number  consists  of  one,  two,  or  three  figures,  its 
cube  root  consists  of  one  figure. 

2.  The  cube  root  of  1000  is  10,  and  the  cube  root  of 
1000000  is  100;  between  1000  and  1000000  are  all  num- 
bers consisting  of  four,  five,  or  six  figures,  and  between 
10  and  100  are  all  numbers  consisting  of  two  figures; 
therefore, 

When  a  number  consists  of  four,  five,  or  six  figures,  its 
cube  root  consists  of  two  figures. 

3.  In  like  manner  it  may  be  shown  that. 

When  a  number  consists  of  seven,  eight,  or   nine  figures^ 
its  cube  root  consists  of  three  figures. 
And  so  on ;  therefore, 

1st.  If  a  number  be  pointed  off  into  periods  of  three  figures 
each,  the  number  of  periods  will  be  the  same  as  the  number 
of  figures  in  the  cube  root. 

2d.  The  cube  of  the  units  will  he  found  in  the  first  period, 
the  cube  of  the  tens  in  the  second  period,  the  cube  of  the 
hundreds  in  the  third  period,  etc. 

243.  To  point  off  a  number  into  periods  of  three 
figures  each. 

1.  Point  off  876453921.  87^453921. 

2.  Point  off  7.356849227.  7.356849227 

3.  Point  off  37683.5624.  37683.562400 

Rule. — Place  a  point  over  the  order  units,  and  then  over 
every  third  order  from  units  to  the  left  and  to  the  right. 

Rem.  1. — The  first  period  on  the  left  of  the  integral  part  of  the 
number  will  often  contain  but  one  or  two  figures. 

Rem.  2. — When  the  first  period  on  the  right  of  the  decimal  part 
contains  but  one  or  two  figures,  ciphers  must  be  annexed  to  complete 
the  period. 

4.  Point  off^  138975462;    3.561325482;  684536.256403. 


CUBE  KOOT. 


311 


5.  Point  off  2756.56843 ;    98451.3276;    .856375;    .0064. 
244.     To  extract  the  cube  root  of  a  number. 
1.  Extract  the  cube  root  of  13824. 


OPERATION. 


13824(24 


2X2X300=:120() 
2X4X  30=  240 
4X4  ■=       10 


1456 


i824 


5824 


Solution.  —  Point  off 
13824  into  periods  of 
three  figures  each  by 
placing  a  point  over  4 
unci   3    (Art.  243,  Rule). 

The  largest  cube  in  13 
(Art.  235,  9,  Table)  is  8; 
its  root  is  2;  place  the 
root  2  on  the  right,  and 
subtract  the  cube  8  from 
13;  the  remainder  is  5, 
to  which  bring  down  the  next  period  824. 

Square  the  root  2  and  multiply  it  by  300;  the  result,  1200,  is  the 
trial  divisor.  Find  how  many  times  1200  is  contained  in  5824;  the 
result  is  4;  place  4  in  the  root  on  the  right  of  2. 

Multiply  2  b\^  4  and  by  30, and  square  4;  add  the  products  240  and 
IG  to  1200;  the  sum  1456  is  the  complete  divisor.  Multiply  1456  by 
4,  and  subtract  the  product  5824  from  5824;  the  remainder  is  0. 
Therefore,  13824  is  a  perfect  cube,  and  its  cube  root  is  24. 


GEOMETRICAL  EXPLANATION. 


After  finding  that  the  cube  root  of 
the  given  number  will  contain  two 
places  of  figures  (tens  and  units),  and 
that  the  figure  in  the  tens'  place  is 
2,  form  a  cube,  A,  Fig.  1,  20  (2  tens) 
inches  long,  20  in.  wide,  and  20  in. 
high;  this  cube  will  contain,  (Art. 
70,)  20X20X20  =  8000  cu.  in.; 
take  this  sum  from  the  whole  number 
of  cubes,  and  5824  cu.  in.  are  left, 
which  correspond  to  the  number  5824 
in  the  numerical  operation. 


312 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


It  is  obvious  that  to  increase  the  figure  A,  and  at  the  sarr.e  timo 
preserve  it  a  cube,  the  length,  breadth,  and  height  must  each  receive 
an  equal  addition.  Then,  since  each  side  is  20  in.  long,  square  20, 
which  gives  20  X  20  =  400,  for  the  number  of  square  inches  in  each 
face  of  the  cube;  and  since  an  addition  is  to  be  made  to  three 
sides,  multiply  the  400  by  3,  which  gives  1200  for  the  number  of 
square  inches  in  the  3  sides.  This  1200  is  called  the  trial  divisor; 
because,  by  means  of  it,  the  thickness  of  the  additions  is  determined. 

By  examining  Fig.  2  it  will  be  seen  that,  after  increasing  each  of 
the  three  sides  equally,  there  will  be  required  3  oblong  solids,  C,  C,  C, 
of  the  same  length  as  each-  of  the  sides,  and  whose  thickness  and 
height  are  each  the  same  as  the  additional  thickness;  and  also  a  cube, 
D,  whose  length,  breadth,  and  height  are  each  the  same  as  the 
additional  thickness.  Hence,  the  solid  contents  of  the  first  three 
rectangular  solids,  the  three  oblong  solids,  and  the  small  cube,  must 
together  be  equal  to  the  remainder  (5824). 

Now  find  the  thickness  of  the 
additions.  It  will  always  be 
.  something  less  than  the  number 
of  times  the  trial  divisor  (1200) 
is  contained  in  the  dividend 
(5824).  By  trial,  we  find  1200 
is  contained  4  times  in  5824; 
proceed  to  find  the  contents  of 
the  different  solids.  The  solid 
contents  of  the  first  three  addi- 
tions, B,  B,  B,  are  found  by  mul- 
tiplying the  number  of  sq.  in.  in 
the  face  by  the  thickness  (Art. 
70);  there  are  400  sq.  in.  in  the 
face  of  each,  and  400  X  S  —  1200 
sq.  in.  in  one  face  of  the  three;  then,  multiplying  by  4  (the  thick- 
ness) gives  4800  cu.  in.  for  their  contents.  The  solid  contents  of  the 
three  oblong  solids,  C,  C,  C,  are  found  by  multiplying  the  number 
of  sq.  in.  in  the  face  by  the  thickness;  now  there  are  20  X  4  =  80  sq. 
in.  in  one  face  of  each,  and  80  X  3  =  240  sq.  in.  in  one  face  of  the 
three;  then  multiplying  by  4  (the  thickness),  gives  960  cu.  in.  for 
their  contents.  Lastly,  find  the  contents  of  the  small  cube,  D,  by 
multiplying  together  its  length,  breadth,  and  thickness;  this  gives 
4X4X4=-  04  cu.  in. 


Fig.  2. 


CUBE  ROOT. 


313 


If  the  solid  contents  of  the  several 
additions  be  added  together,  as  in 
the  margin,  their  sum,  5824  cu.  in. 
will  be  the  number  of  small  cubes 
remaining  after  forming  the  first  cube, 
A.     Hence,   when    13824    cu.   in.    are 

arranged  in  the  form  of  a  cube,  each  side  is  24  in.;  that  is,  the  cube 
root  of  13824  is  24. 


ADDITIONS. 

B  B  Br=4800  cu.  in. 

C  C  C=:    960    "     " 

D=       64    "      " 

Sum,  5  8  2  4 


Kem. — It  is  obvious  that  the  additions  in  the  margin  may  readily 
be  arranged  in  the  same  way  as  in  the  operation  of  the  example. 

2.  Extract  the  cube  root  of  413.5147. 


OPERATION. 


413.5  14  700(7.4  5  + 
343 


7X     7X300r^       14700 

7X     4X     30=  840 

4X4  =  16 


15556 


74X74X300  =  1642800 

74X     SX     3  0=        11100 

5X     5  25 


1653925 


70514 


62224 


8290700 


8269625 


2  10  7  5 


Solution.— Point  off  413.5147  into  periods  of  three  figures  each  by 
placing  a  point  over  3,  and  then  over  4  and  0  to  the  right  (Art.  243, 
Rule).  Then  find  the  figures  of  the  root  as  in  Ex.  1.  The  last  nv 
mainder  is  21075.  Therefore,  413.5147  is  an  imperfect  cube,  and  its 
cube  root  is  7.45 -|-. 

Kem. — By  bringing  down  one  or  more  periods  of  decimal  ciphers 
the  operation  might  be  continued  to  any  required  number  of  decimal 
places  in  the  root. 


3.  Extract  the  cube  root  of  yVg¥?- 


314  KAY'S  NEW  PRACTICAL  AKITIIMETIC. 

Solution. — The  cube  root  of  the  numerator  2197  is  13  and  the 
cube  root  of  the  denominator  13824  is  24;  (Kx.  1);  then,  the  cube 

root  of  tVbV?  i«  \l- 

4.  Extract  the  cube  root  of  ^. 

Solution. — i  reduced  to  a  decimal  is  .8.  The  cube  root  of  .8  to 
three  decimal  places  is  .928;  (Ex.  2);  then,  the  cube  root  of  ^ 
is  .928  +. 

Bule. — 1.  Point  off  the  given  number  into  periods  of  three 
figures  each. 

2.  Find  the  greatest  cube  in  the  first  period  on  the  left; 
place  its  root  on  the  right,  like  a  quotient  in  division ;  sub- 
tract the  cube  frojn  the  period,  and  to  the  remainder  bring 
down  the  next  period  for  a  dividend. 

3.  Square  the  root  found,  and  multiply  it  by  300  for  a 
trial  divisor.  Find  how  many  times  the  trial  divisor  is  con- 
tained, in  the  dividend,  and  place  the  quotient  in  the  root. 

4.  Midfiply  the  preceding  figure,  or  figures,  of  the  root  by 
the  last  and  by  30,  and  square  the  last  figure;  add  the 
products  to  the  trial  divisor ;  the  sum  is  the  complete  divisor. 

5.  Multiply  the  complete  divisor  by  the  last  figure  of  the 
root;  subtract  the  product  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

6.  Find  a  new  trial  divisor  as  before,  and  continue  the 
operation  in  the  same  manner  until  all  the  periods  are 
brought  down. 

Rem.  1. — When  the  number  is  an  imperfect  cube,  the  operation 
may  be  continued  to  any  required  number  of  decimal  places  in  the 
root  by  bringing  down  periods  of  decimal  ciphers.     (Ex.  2). 

Rem.  2. — To  extract  the  cube  root  of  a  common  fraction:  (1)  when 
both  terms  are  perfect  cubes,  extract  the  cube  root  of  the  numerator 
and  then  of  the  denominator;  (Ex.  3);  when  both  terms  are  not  per- 
fect cubes,  reduce  the  fraction  to  a  decimal  and  extract  the  cubo 
root  of  the  decimal.     (Ex.  4). 


CUBE  ROOT. 


315 


Extract  the  cube  root  of 


5. 

91125. 

45. 

15. 

53.1573^ 

16. 

3.76 

6. 

195112. 

58. 

16. 

.199176704. 

.584 

7. 

8. 

9. 

10. 

912673. 
1225043. 
13312053, 
102503232. 

97. 
107. 
237. 

468. 

17. 
18. 
19. 
20. 

tWAV 

if 
If. 
If- 

11. 

529475129. 

809. 

21. 

2. 

1.259+ 

12. 

958585256. 

986. 

22. 

9. 

2.080+ 

13. 

14760213677. 

2453. 

23. 

200. 

5.848+ 

14. 

128100283921. 

5041. 

24. 

91 

2.092+ 

245.  Given  the  solid  contents  of  a  cube  to  find  its 
side  (Art.  70). 

Rule. — Extract  the  cube  root  of  the  solid  contents. 

1.  The  contents  of  a  cubical  cellar  are  1953.125  cu.  ft.: 
find  the  length  of  one  side.  12.5  ft. 

2.  Sixtj^-four  3-incli  cubes  are  piled  in  the  form  of  a 
cube:  what  is  the  length  of  each  side?  1  ft. 

3.  A  cubical  box  contains  512  half-inch  cubes :  what 
are  the  dimensions  of  the  box  inside?  4  in. 

4.  A  cubical  excavation  contains  450  cu.  yd.  17  cu.  ft. : 
w^hat  are  its  dimensions?  23  ft. 

5.  Find  the  side  of  a  cube  equal  to  a  mass  288  ft. 
long,  216  ft.  broad,  and  48  ft.  high.  144  ft. 

6-  The  side  of  a  cubical  vessel  is  1  foot:  find  the  side 
of  another  cubical  vessel  that  shall  contain  3  times  as 
much.  17.306+  in. 


X  MENSURATION.)! 


I.  MEASUREMENT  OF  SURFACES. 


DEFINITIONS. 

246.     1.  A  line  has  length  without  breadth  or  thick- 
ness. 

2.  Lines  are  either  straight  or  curved.     ^^^^  ~^^- 

3.  When  two  lines  meet,  the}'^  form 
an  angle. 

Rem. — The   point   at  which  the  lines  meet   is   called   the   vortex 
of  the  angle. 


4.  Angles  are  either  acute.,  obtuse^  or 
right  angles. 

5.  When  two  straight  lines  are  perpen- 
dictular  to  each  other,  they  form  a  right 
angle. 

6.  An  acute  angle  is  less  than  a  right 
angle. 

7.  An  obtuse  angle  is  greater  than  a 
right  angle. 

8.  When  tw^o  straight  lines  are  ever^^- 
where  equally  distant  they  are  parallel. 

9.  A  surface  has  length  and  breadth 
without  thickness. 

(316) 


MENSURATION.  317 

10.  Surfaces  arc  either  plane  or  curved. 

Thus,  the  surface  of  a  table  or  floor  is  plane;  that  of  a  ball  or 
g^lobe  is  curved. 

11.  A  plane  figure  is  a  portion  of  a  plxme  surface 
bounded  by  one  or  more  lines. 

12.  A  polygon  is  a  plane  figure  bounded  by  straight 
lines. 

Rem. — The  straight  lines  are  called  the  sides  of  the  polygon;  the 
perimeter  of  a  polygon  is  the  sum  of  all  its  sides. 

13.  A  triangle    is    a    plane    figure  /1\ 
bounded    by  three    straight    lines.                  /       I     \^ 

Rem. — If  one  side  be  taken  for  the  base,  the  perpendicular  let 
fall  upon  the  base  from  the  opposite  angle  is  called  the  altitude 
of  the  triangle. 

14.  A  quadrilateral  is  a  plane  figure  bounded  by  four 
straight  lines. 

15.  There   are   three   kinds   of  quadrilaterals :   the  tra- 
pezium^ the  trapezoid^  and  the  parallelo- 
gram. 

16.  A   trapezium    is    a    quadrilateral 
v;ith    no    two   sides   parallel. 

17.  A  trapezoid  is  a  quadrilateral  /  I 
with  only  two  sides  parallel.                               / 

18.  A    parallelogram    is     a     quadri- 

lateral    with    its    opposite     sides    equal       /  / 

and    parallel.  /  / 

Rem.— If  one  side  be  taken  as  the  base,  the  perpendicular  let 
fall  upon  the  base  from  the  opposite  side  is  called  the  altitude  of  the 
parallelogram. 


318 


KAYS   NEW  PKACTICAL  ARITHMETIC. 


19.  A  rhombus  is  a  parallelogram  with  all  its  sides 
equal,  and  its  angles  not  right  angles. 

20.  A  rectangle  is  a  parallelogram  with 
all  its  angles  right  angles. 

21.  A  square  is  a  rectangle  with  all 
its  sides  equal. 

22.  A  polygon  of  five  sides  is  called  a 
pentagon ;  of  six,  a  hexagon ;  of  eight,  an 
octagon,  etc. 

23.  A  diagonal  is  a  line  joining  two 
angles  not  adjacent. 

24.  A  circle  is  a. plane  figure  hounded 
hy  a  curved  line,  every  point  of  which 
is  equally  distant  from  a  point  within 
called  the  center. 

25.  The  circumference  of  a  circle  is 
the  curved  line  which  bounds  the  figure. 

26.  The  diameter  of  a  circle  is  a  straight  line  pass- 
ing through  the  center,  and  terminated,  both  Avays,  by 
the    circumference. 

27.  The  radius  of  a  circle  is  a  straight  line  drawn 
from  the  center  to  the  circumference;  it  is  half  the 
diameter. 


247.     To  find  the  area    of    a    parallelogram  (Art.  246, 
18,  19,  20,  21). 

Rule. — Multiply  the  base  by  the  altitude. 

Explanation. — The  area  of  a  paral- 
lelogram is  equal  to  the  area  of  a  rect- 
angle, having  an  equal  base  and  the  same 
altitude;  but  the  area  of  the  rectangle  is 
equal    to    its    length    multiplied    by    its 

breadth;   (Art.  68);   hence,  the  area  of  a  parallelogram  is  equal  to 
its  base  multiplied  by  its  altitude. 


MENSURATION.  319 

1.  How  many  square  feet  in  a  floor  17  ft.  long  and 
15  ft.  Avide?  255  sq.  ft. 

2.  How  many  acres  of  land  in  a  parallelogram,  the 
length  of  which  is  120  rd.,  and  the  perpendicular  breadth 
84  rd.  ?  63  A. 

3.  How  many  acres  in  a  square  field,  each  side  of 
which  is  65  rd.  ?  26  A.  65  sq.  rd. 

4.  How  many  acres  in  a  field  in  the  form  of  a  rhom- 
bus, each  side  measuring  35  rd.,  and  the  perpendicular 
distance  between  two  sides  being  16  rd.  ? 

3  A.   80  sq.  rd. 

5.  Find  the  difference  in  area  between  a  floor  30  ft. 
square,  and  two  others  each   15  ft.  square.  50  sq.  yd- 

6.  A  table  is  3  ft.  4  in.  long,  and  2  ft.  10  in.  wide: 
how  many  sq.  ft.  in  its  surface? 

Solution.— 3  ft.  4  in.r=3J  or  -i/-  ft.;  2  ft.  10  in.  =  2|  or  -i/  ft.; 
then,  the  surface  of  the  table  is  }^oy^lJ_  —  ^  gq,  ft.     Qr 

Solution. — 3  ft.  4  in.:r=40  in.;  2  ft.  10  in.  =  34  in.;  then  the  sur- 
face of  the  table  is  40X34^=1360  sq.  in.;  1360 -- 144=^9  sq.  ft. 
64  sq.  in.,  or  9|  sq.  ft. 

7.  How  many  square  feet  in  a  marble  slab  5  ft.  6  in, 
long  and  1  ft.  8  in.  wide?  dj-  sq.  ft. 

8.  How  many  square  yards  in  a  ceiling  25  ft.  9  in, 
long,  and  21  ft.  3  in.  wide? 

60  sq.  yd.  7  sq.  ft.  27  sq.  in. 

9.  A  room  is  10  ft.  long:  how  wide  must  it  be  to 
contain  80  sq.  ft.  ?  8  ft. 

10.  How  many  yards  of  carpet,  1^  yd.  wide,  will  cover 
a  floor  18  ft.  long  and  15  ft.  wide?  20  yd. 

11.  How  many  yards  of  flannel,  |  yd.  wide,  will  it 
take  to  line  3  yd.  of  cloth,  li^  yd.  wide?  6  yd. 


320  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

12.  How  mtiny  yards  of  carpet,  IJ  yd.  wide,  will  it 
lake  to  cover  a  floor  21  ft.  8  in.  long  and  13  ft.  G  in. 
wide?  25iyd. 

13.  A  rectangular  field  is  15  rd.  long:  what  must  be 
its  width  to  contain   1  A.  ?  lOf  rd. 

248.     To  find  the  area  of  a  trapezoid  (Art.  246,  17). 

Rule  — Multiply  half  the  sum  of  the  jxiralld  sides  by 
the  altitude. 


Explanation. — The  base  of  a  prirallclo- 
grarn  having  the  same  altitude  and  equal 
area  is  one-half  the  sum  of  the  parallel 
sides  of  the  trapezoid. 

1.  The  parallel  sides  of  a  trapezoid  are  2  ft.  2  in.  and 
2  ft.  11  in.;  its  altitude  is  11  in.:  what  is  its  area? 

2  sq.  ft.  47^  sq.  in. 

2.  A  field  is  in  the  form  of  a  trapezoid  ;  one  of  the 
parallel  sides  is  25  rd.,  and  the  other  19  rd. ;  the  width 
is  32  rd. :  how  many  acres  in  the  field? 

4  A.  64  sq.  rd. 

3.  How  many  square  yards  in  a  piece  of  roof  10  ft.  8 
in.  wide  on  the  lower  side,  and  6  ft.  2  in.  wide  on  the 
upper  side,  the  length  being  12  ft.?      11  sq.  yd.  2  sq.  ft. 

249.     To  find  the  area  of  a  triangle. 

Ist.  When  the  base  and  altitude  are  given. 

Rule. — Multiply  the  base  by  the  altitude^  and  take  half 
the  product. 

Explanation.— The  area  of  a  triangle                 /\  7 

is  one-half  the  area  of  a   parallelogram               /           \^       / 
having  the  same  base  and  altitude.  /        I _\^/ 


MENSUKATION.  321 

2d.  When  the  three  sides  are  given. 

Rule. — 1.  From  half  the  sum  of  the  three  sides  take 
each  side  separately, 

2.  Midtiply  the  half-sum  and  the  three  remainders  to- 
gether ^  and   extract   the   square   root   of  the  product. 

1.  The  base  of  a  triangle  is  15  ft.  and  its  altitude  12 
ft. :  what  is  its  area  ?  90  sq.  ft. 

2.  One  side  of  a  triangular  lot  is  44  rd.,  and  the  per- 
pendicular distance  from  the  angle  opposite  to  this  side 
is  18  rd. :  how  many  acres  in  the  lot?       2  A.  76  sq.  rd. 

3.  What  is  the  area  of  a  triangle,  of  which  the  base 
is  12  ft.  6  in.  and  the  altitude  16  ft.  9  in.? 

11  sq.  yd.  5  sq.  ft,  99  sq.  in. 

4.  Find  the  area  of  a  triangle  whose  sides  are  13,  14, 
and  15  ft.  84  sq.  ft. 

5.  The  sides  of  a  triangle  are  30,  40,  and  50  ft. :  what 
is  the  area?  QQ  sq.  yd.  6  sq.  ft. 

250.  To  find  the  area  of  a  trapezium  (Art.  246,  16) 
or.  other  irregular  figure. 

Rule. — 1.  Divide  the  figure  into  triangles  by  diagonals. 
2.  Find  the  areas  of  the  triangles,  and  add.  them  together. 

1.  Find  the  area  of  a  field  in  the  form  of  a  trapezium, 
of  which  a  diagonal  is  50  rd.  and  the  perpendiculars  to 
the  diagonal  from  the  opposite  angles  30  rd.  and  20  rd. 

7  A.  130  sq.  rd. 

251.  1.  To  find  the  circumference  of  a  circle  when  the 
diameter  is  given. 

Rule. — Multiply  the  diameter  by  3.1416. 

Prac.  21, 


322  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

2.  Conversely :  to  find  the  diameter  of  a  circle  when 
the  circumference  is  given. 

Rule. — Divide  the  circumference  by  3.141G. 

1.  The  diameter  of  a  circle  is  48  fl. :  what  is  the  cir- 
cumference? 150  fl.  9.56  in. 

2.  The  circumference  of  a  circle  is  15  fl. :  what  is  the 
diameter?  4  fl.  9.3  in. 

3.  The  diameter  of  a  wheel  is  4  fl. :  what  is  its  cir- 
cumference? 12  fl.  6.8  in. 

4.  If  the  girth  of  a  tree  is  12  fl.  5  in.,  what  is  its 
diameter?  3  fl.  11.43  in. 

5.  What  is  the  circumference  of  the  earth,  the  diam- 
eter being  7912  mi.?  24856+  mi. 

252.  1.  To  find  the  area  of  a  circle,  when  the  radius 
is  given. 

Rule. — Multiply  the  square  of  the  radius  by  3.1416. 

2.  Conversely :  to  find  the  radius  of  a  (drcle  when  the 
area  is  given. 

Rule. — Divide  the  area  by  3.1416,  and  extract  the  square 
root  of  the  quotient. 

1.  Find  the  area  of  a  circle  whose  radius  is  21  fl. 

153  sq.  yd.  8  sq.  ft.  64  sq.  in. 

2.  The  area  of  a  circle  is  6  sq.  fl.  98.115  sq.  in. :  what 
are  its  diameter  and  circumference? 

2  ft.  11  in.;  9  fl.  1.9+  in. 

3.  How  long  a  rope  will  it  take  to  fasten  a  horse  to 
a  post  so  that  he  may  graze  over  1  A.  of  grass,  and  no 
more  ?  7  rd.  2  ft.  3  in. 


MEASUREMENT  OF  SOLIDS.  323 

4.  Two  circles,  10  and  16  ft.  in  diameter,  have  the 
same  center :  what  is  the  area  of  the  ring  between  their 
circumferences?  122  sq.  ft.  75  sq.  in. 

5.  The  area  of  a  circle  is  1  square  foot,  what  is  its 
diameter?  13.54  in. 


II.   MEASUREMENT  OF  SOLIDS. 
DEFINITIONS. 

253.     1.  A  solid,   or  body,  has   length,   breadth,   and 
thickness. 

2.  A  prism  is  a  solid  with  two  parallel  bases, 
which  are  ^^olygons,  and  with  its  faces  parallel- 
ograms. 

Rem. — A  prism  is  triangular,  quadrangular,  etc.,  ac- 
cording to  the  shape  of  the  base. 

3.  A  right  prism  has  its  faces  rectangles. 

4.  The  altitude   of  a   prism    is    the   perpendicular   let 
fall  from  one  base  upon  the  other. 

5.  The  convex  surface  of  a  prism  is  the  sum  of  the 
areas  of  its  faces. 

6.  A  parallelopipedon  is  a  prism  with  its  bases  paral- 
lelograms. 

7.  A  right  parallelopipedon  is  a 
solid  with  six  rectangular  faces  (Art. 
70). 

8.  A    cube    is    a    solid    with    six    equal 
square   faces. 

9.  A  pyramid  is  a   solid  with    one  base, 
which  is  a  polygon,  and  with  its  faces  triangles. 


324 


RAY'S  NEW  PRACTICAL  ARITHMETIC. 


10.  A  right  pyramid  has  all  its  faces  equal. 

11.  The  slant  height  of  a  right  pyramid  is 
the  distance  from  the  vertex  to  the  middle  of 
each  side  of  the  base. 

12.  The  three  round  bodies  are  the  cylinder^ 
the  cone^  and  the  sphere. 

13.  A  cylinder  is  a  solid  with  two  paral- 
lel bases,  which  are  circles,  and  with  a 
curved  surface. 

14.  The  axis  oi^  a  cylinder  is  a  line  join- 
ing the  centers  of  the  two  bases. 

15.  The  convex  surface  of  a  cylinder  is 
the  area  of  its  curved  surface. 

16.  A  cone  is  a  solid  with  one  base, 
which  is  a  circle,  and  with  a  curved 
surface  terminating  in  an  apex. 

17.  A  sphere  is  a  solid  with  a 
curved  surface,  every  point  of  which 
is  equally  distant  from  a  point 
within    called    the    center. 

18.  The  volume  of  a  body  is  its 
solid   contents. 


254.     1.  To  find  the  convex  surface  of  a  right  prism. 

Rule. — Multiply  the  perimeter  of  the  base  by  the  altitude, 

2.  To  find  the  convex  surface  of  a  cjdinder. 

Rule. — Multiply    the    circumference    of  the    base    by    the 
altitude. 


3.  To    find    the    entire    surface    of   a    prism,    or    of   a 
cylinder. 


MEASUREMENT  OF  SOLIDS.  325 

Rule. —  To  the  convex  surface  add  the  areas  of  the  two 
bases. 

1.  Find    the    surface  of  a  cube,  each  side  being  37  in. 

6  sq.  yd.  3  sq.  ft.  6  sq.  in. 

2.  Find  the  surface  of  a  right  prism,  with  a  trian- 
gular base,  each  side  of  whicii  is  4  ft. ;  the  altitude  of 
the  prism  is  5  ft.  73.85  -f-  sq.  ft. 

3.  Find  the  surface  of  a  box  which  is  3  ft.  6  in.  long, 
2  ft  9  in.  wide,  and  1  ft.  10  in.  high..  42^  sq.  ft. 

4.  Find  the  surface  of  a  cylinder,  its  altitude  being  5 . 
ft.  and  the  radius  of  the  base,  2  ft.  87.96  +  sq.  ft. 

255.     To  find  the  volume  of  a  prism  or  of  a  cylinder. 

Rule. — Multiply  the  area    of  the   base   by   the   altitude. 

Rem. — The  rule  for  finding  the  volume  of  a  right  parallelopipedon 
is  given  in  Art.  70. 

1.  Find    the    volume    of   a    right    parallelopipedon,   of 
which  the  length  is  12  ft.,  the  width  3  ft.  3  in.,  and  the 
4  ft.  4  in.  169  cu.  ft. 


SoLTJTiON.— 3  ft.  3  in.  =  3|:  or  -i^3  ft.;  4  ft.  4  m.  =  ^  or  -i/  ft.; 
then,  the  volume  of  the  parallelopipedon  is  12  X -V"  XV'^ -^^^ 
cu.  ft. 

2.  How  many  cubic  yards  in  a  room  24  ft.  long,  18 
ft.  6  in.  wide,  and   10  ft.  7  in.  high? 

174  cu.  yd.  1  cu.  ft. 

3.  Each  side  of  the  base  of  a  triangular  prism  is  2  ft.; 
its  altitude  is  14  ft.:  what  is  the  volume  of  the  prism? 

24J  cu.  ft.  nearly. 

4.  Find  the  volume  of  a  cylinder  whose  altitude  is  12 
ft.  and  the  radius  of  the  base  2  ft.  150.8  cu.  ft.  nearly. 


326  RAY'S  NEW   PRACTICAL  ARITHMETIC. 

5.  How  many  cubic  inches  in  a  peck  measure,  the 
diameter  of  the  bottom  being  9J  in.  and  the  depth  8  in.  ? 

537.6  -f  cu.  in. 

256.  1.  To  find  the  convex  surface  of  a  right 
pyramid. 

Rule. — Multiply  the  perimeter  of  the  base  by  the  slant 
height^  and  take  half  the  product. 

2.  To  find  the  convex  surface  of  a  cone. 

Rule. — Multiply  the  circumference  of  the  base  by  the  slant 
height^  and  take  half  the  jnoduct. 

3.  To  find  the  entire  surface  of  a  p3'ramid  or  of  a 
cjone. 

Rule. —  To  the  convex  surface  add  the  area  of  the  base. 

1.  Find  the  entire  surface  of  a  right  pyramid,  with  a 
triangular  base,  each  side  of  which  is  5  ft.  4  in. ;  the 
slant  height  of  the  pyramid  is  7  fl.  6  in.      72.3 -f  sq.  ft. 

2.  What  is  the  convex  surface  of  a  cone  of  which  the 
slant  height  is  25  ft.  and  the  diameter  of  the  base  8  ft. 
6  in.?  333.8  sq.  ft.  nearly. 

3.  Find  the  entire  surface  of  a  cone,  of  which  the  slant 
height  is  4  ft.  7  in.  and  the  diameter  of  the  base  2  ft. 
11  in.  27.6 +  sq.  ft. 

257.  To  find   the  volume  of  a  pyramid  or  of  a  cone. 

Rule. — Multiply  the  area  of  the  base  by  the  altitude^ 
and  take  one-third  of  the  product. 

1.  Find  the  volume  of  a  square  pyramid,  of  which 
each  side  of  the  base  is  5  ft.  and  the  altitude  21  ft. 

175  cu.  ft. 


MEASUREMENT  OF  SOLIDS.  327 

2.  Find  the  volume  of  a  cone,  of  which  the  altitude  is 
15  ft.  and  the  radius  of  the  base  5  ft.  392.7  cu.  ft. 

3.-  A  square  pyramid  is  477  ft.  high  ;  each  side  of  its 
base  is  720  ft. :  how  many  cubic  yards  in  the  pyramid  ? 

3052800  cu.  yd. 

4.  The  diameter  of  the  base  of  a  conical,  glass  house,  is 
37  ft.  8  in.,  and  its  altitude  79  ft.  9  in. :  what  is  the 
space  inclosed?  29622  +  cu.  ft. 


258.  To  find  the  surface  of  a  sphere. 

Rule. — Multiply  the  square  of  the  diameter  by  3.1416. 

1.  What  is  the  surface  of  a  sphere,  of  which  the  diam- 
eter is  1ft.?  3.14  +  sq.  ft. 

2.  What  is  the  surface  of  a  sphere,  of  which  the  diam- 
eter is  4  ft.  6  in.  ?  63.6  +  sq.  ft. 

3.  What  is  the  area  of  the  earth's  surface,  on  the 
supposition  that  it  be  a  perfect  sphere  7912  miles  in 
diameter?  196663355.75  +  sq.  mi. 

259.  To  find  the  volume  o-f  a  sphere. 

Bule. — Multiply  the  cube  of  the  diameter  by  one-sixth 
of  3.1416,  or  .5236. 

1.  Find  the  volume  of  a  sphere  13  ft.  in  diameter. 

1150.3 +  CU.  ft. 

2.  Find  the  volume  of  a  sphere  2  ft.  6  in.  in  diam- 
eter. 8.2  cu.  ft.  nearly. 

3.  The  volume  of  a  sphere  is  1  cu.  ft. :  what  is  its 
diameter?  14.9  in.  nearly. 


328  UAY\S  NEW  PRACTICAL  AKITHMETIC. 

III.     APPLICATIONS    OF    MENSURATION. 

260.  1.  Plastering,  house-painting,  paving,  paper- 
hanging,  etc.,  are  meaBured  by  the  square  foot  or  square 
yard. 

2.  Glazing  is  measured  by  the  square  foot  or  by  the 
pane.  ' 

3.  Stone  cutting  is  measured  by  the  square  foot. 

4.  Flooring,  roofing,  etc.,  are  measured  by  the  square 
yard  or  by  the  square  of  100  sq.  ft. 

1.  A  room  is  20  ft.  6  in.  long,  16  fl.  3  in.  broad,  10 
ft.  1  in.  high :  how  many  yards  of  plastering  in  it,  de- 
ducting a  fire-place  6  ft.  3  in.  by  4  ft.  2  in.  ;  a  door  7 
ft.  by  4  ft.  2  in.,  and    two   windows,  each  6  ft.  b}'  3. ft. 

3  in.  ?  108  sq.  yd.  8  sq.  ft.  6  sq.  in. 

2.  A  room  is  20  ft.  long,  14  ft.  6  in.  broad,  and  10 
ft.  4  in.  high:  what  will  the  papering  of  the  walls  cost, 
at  27  ct.  per  square  3'ard,  deducting  a  fire-place  4  ft.  by 

4  ft.  4  in.,  and  two  windows,  each  6  ft.  by  3  ft.  2  in.? 

SI  9.73. 

3.  What  will  it  cost  to  pave  a  rectangular  court,  21 
yd.  long  and^VlS  j&.  broad,  in  which  a  foot-path,  5  ft. 
wide,  runs  the  whole  length:  the  path  paved  with  flags, 
at  36  ct.  per  square  3'ard,  and  the  rest  with  bricks,  at 
24  ct.  per  square  yard  ?  ^79.80. 

4.  At  10  ct.  a  square  yard,  what  will  it  cost  to  paint 
both  sides  of  a  partition  15  ft.  6  in.  long,  12  ft.  6  in. 
high?  S4.31. 

5.  A  house  has  three  tiers  of  window^s,  seven  in  a 
tier:  the  height  of  the  first  tier  is  6  ft.  11  in.;  of  the 
second,  5  ft.  4  in. ;  of  the  third,  4  ft.  3  in. ;  each  win- 
dow is  3  ft.  6  in.  wide:  what  will  the  glazing  cost,  at 
16  ct.  per  square  foot?  $64.68. 


APPLICATIONS  OF  MENSUKATION.  329 

6.  A  floor  is  36  ft.  3  in.  long,  16  ft.  6  in.  wide :  what 
will  it  cost  to  lay  it,  at  $3  a  square?  $17.94. 

7.  At  S3.50  jDer  square,  what  will  be  the  cost  of  a 
roof  40  ft.  long,  the  rafters  on  each  "  side  18  ft.  6  in. 
long?  $51.80. 

BOARD    MEASURE. 

261,  1.  Board  Measure  is  used  in  measuring  all 
lumber  which  is  sawed  into  boards,  planks,  etc. 

2.  A  foot,  board  measure,  is  1  foot  long,  1  foot  wide, 
and  1  inch  thick. 

3.  Hence,  to  find  the  number  of  feet  in  a  board,  we 
have  the  following 

Rule. — 1.  Find  the  surface  of  the  hoard  in  square  feet. 
2.  Multiply  the  surface  by  the  thickness  in  inches. 

1.  How  many  feet  in  an  inch  board  16  ft.  long  and  1 
ft.  3  in.  wide?  20  ft. 

2.  How  many  feet  in  a  two-inch  plank  12  ft.  6  in. 
long  and  2  ft.  3  in.  wide?  56 J  ft. 

3.  How  many  feet  in  a  piece  of  scantling  15  ft.  long, 
4  in.  wide,  and  3  in.  thick?  15  ft. 

4.  How  many  feet  of  inch  boards  will  a  stick  of  tim- 
ber 12  ft.  long  and  2  ft.  square  make?  576  ft. 

5.  How  many  feet  in  an  inch  board,  12  ft.  6  in.  long 
1  ft.  3  in.  wide,  at  one  end,  and  11  in.  wide  at  the 
other  end?  13i|  fl. 

MASONS'    AND    BRICKLAYERS'    WORK. 

262.  1.  Stone  masonry  is  usually  measured  by  the 
perch,  which  is  24f  or  24.75  cu.  ft.   (Art.  70). 

2.  Bricklaying    is    commonly    measured    by    the    1000 

bricks. 


330  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

1.  How  many  perches  in  a  stone  wall  97  ft.  5  in.  long, 
18  ft.  3  in.  high,  2  ft.  3  in.  thick?  161.6 

2.  What  is  the  cost  of  a  stone  wall  53  ft.  G  in.  long,  12 
ft.  6  in.  high,  2  ft.  thick,  at  ^2.25  a  perch?  $121.59. 

3.  How  many  bricks  in  a  wall  48  ft.  4  in.  long,  16 
ft.  6  in.  high,  1  ft.  6  in.  thick,  allowing  20  bricks  to  the 
cubic  ft.?  23925. 

4.  How  many  bricks,  each  8  in.  long,  4  in.  wide,  2.25 
in.  thick,  will  be  required  for  a  wall  120  ft.  long,  8  ft. 
high,  and  1  ft.  6  in.  thick?  34560. 

5.  Find  the  cost  of  building  a  wall  240  ft.  long,  6  ft. 
high,  3  ft.  thick,  at  $3.25  per  1000,  each  brick  being  9 
in.  long,  4  in.  wide,  and  2  in.  thick.  S336.96. 

MEASUREMENT  BY  BUSHELS  OR  GALLONS. 
263.     1.  To  find  the  number  of  bushels  (Art.  61). 

Rule. — Find  the  volume  in  cubic  inches ,  and  divide  by 
2150.4. 

2.  To  find  the  number  of  gallons  (Art.  64). 

Rule. — Find  the  volume  in  cubic  iiiches,  and  divide  by 
231. 

1.  How  many  bushels  in  a  bin  15  ft.  long,  5  ft.  wide, 
and  4  ft.  deep?  241 +. 

2.  How  many  gallons  in  a  trough  10  ft.  long,  5  ft. 
wide,  and  4  ft.  deep  ?  1496  +. 

3.  How  many  bushels  in  a  cylindincal  tub  6  ft.  in 
diameter  and  8  ft.  deep?  181.76+. 

4.  How  many  barrels,  of  31 J  gal.  each,  in  a  cistern, 
in  the  form  of  a  cylinder,  of  which  the  diameter  is  4  ft. 
and  the  depth  6  ft.?  17.9+  bl. 


I.  ARITHMETICAL   PROGRESSION, 

264.  1.  An  Arithmetical  Progression  is  a  series  of 
numbers  which  increase  or  decrease  by  a  common  difference. 

2.  If  the  series  increase,  it  is  called  an  increasing  series; 
if  it  decrease,  a  decreasing  series. 


Thus,       1,       8,       5,       7, 
20,     17,     14,     11, 


11,  etc.,  is  an  increasing  series. 
5,  etc.,  is  a    decreasing  series. 


3.  The  numbers  forming  the  series  are  called  terms; 
the  first  and  last  terms  are  the  extremes;  the  other  terms, 
the  means. 

4.  In  every  arithmetical  series,  five  things  are  consid- 
ered:  (1)  i\\Q  first  term,  (2)  the  last  term,  (3)  the  com- 
mon difference^  (4)  the  number  of  terms,  and  (5)  the  sum 
of  the  terms. 


CASE    I. 

1     265.     To  find  the  last  term,  when  the  first  term,  the 
common  difference,  and   the  number  of  terms  are  given. 

1.  I  bought  10    yd.  of  muslin,    at  3    ct.    for   the   first 

yard,  7   ct.    for  the  second,  11  ct.  for  the  third,    and    so 

on  :  what  did  the  last  A^ard  cost  ? 

(331) 


332  KAY'S  NEW  PRACTICAL  ARITHMETIC. 

Solution. — To  fihd  the  cost  of  the  second  yard,  operation. 

add  4  ct.  once  to  the  cost  of  the  first;  to  find  the  .  4  X  9  =  36 
cost  of  the  third,  add  4  ct.  twice  to  the  cost  of  the  3  -f  3  6  =  3  9 
first;  to  find  the  cost  of  the  jourth,  add  4  ct. 
three  times  to  the  cost  of  the  first,  and  so  on;  hence,  to  find  the  cost 
of  the  tenth  yard,  add  4  ct.  nine  times  to  the  cost  of  the  first;  but  9 
times  4  ct.  are  36  ct.,  and  3  ct. -f  36  ct.  =  39  ct.,  the  cost  of  the  last 
yard,  or  last  term  of  the  progression. 

2.  The  first  term  of  a  decreasing  scries  is  39,  the  com- 
mon  difference  4,  and  the  number  of  terms  10 :  find  the 
last  term. 

OPERATION. 

Solution. — In  this  case,  4  must  he  subtracted  4  X  9  ==:  3  6 
9  times  from  39,  which  will  give  3  for  the  last  3  9  —  36=  8 
term. 

Rule. — 1.  Multiply  the  common  difference  by  the  number 
of  terms  less  one. 

2.  If  an  increasing  series,  add  the  product  to  the  first 
term;  if  a  decreasing  series,  subtract  the  product  from  the 
first  term. 

3.  Find  the  last  term  of  an  increasing  series  in  which 
the  first  term  is  2,  the  common  difference  3,  and  the 
number  of  terms  50.  149. 

4.  What  is  the  54th  term  of  a  decreasing  series  in  which 
the  first  term  is  140,  and  common  difference  2?  34. 

5.  What  is  the  99th  term  of  a  decreasing  series  in  which 
the  first  term  is  329,  and  common  difference  ^?  243J. 

CASE    II. 

266.  To  find  the  common  difference,  when  the  ex- 
tremes and  the  number  of  terms  are  given. 

1.  The  first  term  of  a  series  is  2,  the  last  20,  and  the 
number  of  terms  7:  what  is  the  common  difference? 


AKITHMETICAL    FROGKESSION.  333 

Solution. — The  difference  of  the  extremes  20  operation. 

and   2   is  18;    18   divided   by  6,  the   number  of        20  —  2  =  18 
terms   less  1,  is  3,  the  common  difference.  18^6=     3 

Rule. — Divide  the  difference  of  the  extremes  by  the  num- 
ber of  terms  less  one. 

2.  The  extremes  are  3  and  300 ;  the  number  of  terms 
10 :  find  the  common  diiferenco.  33. 

3.  A  travels  from  Boston  to  Bangor  in  10  da. ;  he 
goes  5  mi.  the  first  day,  and  increases  the  distance  trav- 
eled each  day  by  the  same  number  of  miles;  on  the  last 
day  he  goes  50  mi. :  find  the  daily  increase.  5  mi. 


CASE   III. 

267.  To  find  the  sum  of  all  the  terms  of  the  series 
when  the  extremes  and  the  number  of  terms  are  given. 

1.  Find  the  sum  of  6  terms  of  the  series  whose  first 
term  is  1,  and  last  term  11. 

Solution. — The  series  is  . 

In  inverted  order  it  is 

The  sum  of  the  two  is     .     .   12^       12^       12^       12^       12;       12. 

Since  the  two  series  are  the  same,  their  sum  is  twice  the  first  series; 
but  their  sum  is  obviously  as  many  times  12,  the  sum  of  the  ex- 
tremes, as  there  are  terms;  hence,  the  sum  of  the  series  is  6  times 
12  =  72  divided  by  2  =  36. 

Rule. — Multiply  the  sum  of  the  extremes  by  the  number 
of  terms;  and  take  half  the  product. 

2.  The  extremes  are  2  and  50 ;  the  number  of  terms, 
24 :  find  the  sum  of  the  series.  624. 

3.  How  many  strokes  does  the  hammer  of  a  clock 
strike  in   12  hours?  78. 


1, 

3, 

5, 

7, 

9, 

11. 

11, 

9, 

7, 

5, 

3, 

1. 

334  RAY'S  NEW  PRACTICAL  ARITHMETIC. 

4.  Place  100  apples  in  a  right  line,  3  yd.  fi-om  each 
other,  the  first,  3  yd.  from  a  basket:  what  distance  will 
a  boy  travel  who  gathers  them  singly  and  places  them 
in  the  basket?  17  mi.  69  rd.  ^  yd. 

5.  A  body  falling  by  its  own  weight,  if  not  resisted 
by  the  air,  would  descend  in  the  first  second  a  space  of 
16  ft.  1  in.;  the  next  second,  3  times  that  space;  the 
third,  5  times  that  space ;  the  fourth,  7  times,  etc. :  at 
that  rate,  through  what  space  would  it  fall  in  1  minute? 

57900  ft. 


II.    OEOMETRICAL    PROGRESSION. 

268.  1.  A  Geometrical  Progression,  is  a  series  of 
numbers  increasing  by  a  common  multiplier^  or  decreasing 
by  a  common  divisor. 

Thus,     1,       3,       9,       27,       81,    is  an  increasing  geometric  series. 
48,     24,     12,         6,         3,    is  a    decreasing  geometric  series. 

2.  The  common  multiplier  or  common  divisor,  is  called 
the  ratio. 

Thus,  in  the  first  of  the  above  series,  the  ratio  is  3;  in  the 
second,  2. 

3.  The  numbers  forming  the  series  are  the  terms;  the 
first  and  last  terms  are  extremes;  the  others,  means. 

4.  In  every  geometric  series,  five  things  are  considered : 
(1)  the  first  term ;  (2)  the  last  term ;  (3)  the  number  of 
terms ;  (4)  the  ratio ;  (5)  the  sum  of  the  terms. 

CASE    I. 

269,  To  find  the  last  term,  when  the  first  term,  the 
ratio,  and  the  number  of  terms  are  giv^en. 


GEOMETRICAL  PROGRESSION.  335 

1.  The  first  term  of  an  increasing  geometric  series,  is 
2 ;  the  ratio,  3 :  what  is  the  fifth  term  ? 

Solution. — The  first  term  is  2;  the  second,  2X3;  the  third, 
2X^X3;  the  fourth,  2X8X3X3;  and  the  fifth,  2X3X3X3X 
3.  Each  term  after  the  first,  consists  of  the  first  term  multiplied  by 
the  ratio  as  many  times  less  one,  as  is  denoted  by  the  number  of  the 
term;  then,  the  fifth  term  consists  of  2  multiplied  by  3  taken  four 
times  as  a  factor;  but  3,  taken  4  times  as  a  factor,  is  the  4tt\\  power 
of  3.     Hence,  the  fifth  term  is  2  X  3^  =.  162. 

2.  The  first  term  of  a  decreasing  geometric  series  is 
192;  the  ratio,  2:  what  is  the  fourth  term? 

Solution. — The  first  term  is  192;  the  second  term  is  192-7-2; 
the  third  is  192  divided  by  2X2;  the  fourth  is  192  divided  by  2  X 
2X2;  that  is,  192  -f-  2=*  =  24. 

Rule. — 1.  Raise  the  ratio  to  a  power  whose  exponent  is 
one  less  than  the  number  of  terms. 

2.  If  the  series  be  increasing,  multiply  the  first  term  by 
this  power;  if  decreasing,  divide  the  first  t^m^by  the  power. 

3.  The  first  term  of  an  increasing  series  is  2 ;  the  ratio, 
2;  the  number  of  terms,  13:  find  the  last  term.        8192. 

4.  The  first  term  of  a  decreasing  series  is  262144 ;  the 
ratio,  4 ;  number  of  terms,  9 :  find  the  last  term.  4. 

5.  The  first  term  of  an  increasing  series  is  10;  the 
ratio,  3:  what  is  the  tenth  term?  196830. 


CASE  II. 

270.     To  find  the  sum  of  all  the  terms  of  a  geometric 
series. 

1.  Find  the  sum    of  5    terms  of  the   geometric  series, 
whose  first  term  is  4,  and  ratio  3. 


336  PvAY'S  NEW  PRACTICAL  ARITHMETIC. 

Solution. — Write  the  terms  of  the  series  as  below;  then  multiply 
each  term  by  the  ratio,  and  remove  the  product  one  term  toward  the 
right,  thus  : 

4  +  12-f36-flO  8  +  324  =:=^  sum  of  the  series. 

12  +  36  +  108  +  324  +  97  2  =r^  sum  X  3. 

Since  the  upper  line  is  once  the  sum  of  the  series,  and  the  lower 
three  times  the  sum,  their  difference  is  twice  the  sum;  hence,  if  the 
upper  line  be  subtracted  from  the  lower,  and  the  remainder  divided 
by  2,  the  quotient  will  be  the  sum  of  the  series.  Performing  this 
operation,  we  have  972  —  4=^968  divided  by  2;  the  quotient  is  484, 
the  sum  of  the  series.  In  this  process,  972  is  the  product  of  the 
greatest  term  of  the  given  series  by  the  ratio,  4  is  the  least  term,  and 
the  divisor  2  is  equal  to  the  ratio  less  one. 

Rule. — Multiply  the  greatest  term  by  the  ratio;  from  the 
product  subtract  the  least  term;  divide  the  remainder  by 
the  ratio  less  1. 

Rem. — When  a  series  is  decreasing,  and  the  number  of  terms  in- 
finite, the  last  term  is  0. 

2.  The  first  term  is  10 ;  the  ratio,  3 ;  the  number  of 
terms,  7:  what  is  the  sum  of  the  series?  10930. 

3.  A  father  gave  his  daughter  on  T^ew  Year's  day  $1; 
he  doubled  it  the  first  day  of  every  month  for  a  year: 
what  sum  did  she  receive?  84095. 

4.  I  sold  1  lb.  of  gold  at  1  ct.  for  the  first  oz. ;  4  ct. 
for  the  second,  16  ct.  for  the  third,  etc. :  what  sum  did 
I  get?  $55924.05. 

5.  Find  the  sum  of  an  infinite  series,  of  which  the 
greatest  term  is  .3  and  the  ratio,  10;  that  is,  of  yu  ~r  toj^ 
+  T0V0'  ^tc.  ^  ^. 

6.  Find  the  sum  of  the  infinite  series  i,  i,  2V?  ^^c.     J. 

7.  Find  the  sum  of  the  infinite  series  i,  ^,  J,  etc.       1. 


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^  14  DA1 

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Ray's  Arithmetics, 
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Eclectic  Geographies, 
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